Calculo I (George B. Thomas) - Resolução dos exercícios
Descrição do Produto
CHAPTER 1 PRELIMINARIES 1.1 REAL NUMBERS AND THE REAL LINE 1. Executing long division,
" 9
2. Executing long division,
" 11
œ 0.1,
2 9
œ 0.2,
œ 0.09,
2 11
3 9
œ 0.3,
œ 0.18,
3 11
8 9
œ 0.8,
œ 0.27,
9 11
9 9
œ 0.9
œ 0.81,
11 11
œ 0.99
3. NT = necessarily true, NNT = Not necessarily true. Given: 2 < x < 6. a) NNT. 5 is a counter example. b) NT. 2 < x < 6 Ê 2 2 < x 2 < 6 2 Ê 0 < x 2 < 2. c) NT. 2 < x < 6 Ê 2/2 < x/2 < 6/2 Ê 1 < x < 3. d) NT. 2 < x < 6 Ê 1/2 > 1/x > 1/6 Ê 1/6 < 1/x < 1/2. e) NT. 2 < x < 6 Ê 1/2 > 1/x > 1/6 Ê 1/6 < 1/x < 1/2 Ê 6(1/6) < 6(1/x) < 6(1/2) Ê 1 < 6/x < 3. f) NT. 2 < x < 6 Ê x < 6 Ê (x 4) < 2 and 2 < x < 6 Ê x > 2 Ê x < 2 Ê x + 4 < 2 Ê (x 4) < 2. The pair of inequalities (x 4) < 2 and (x 4) < 2 Ê | x 4 | < 2. g) NT. 2 < x < 6 Ê 2 > x > 6 Ê 6 < x < 2. But 2 < 2. So 6 < x < 2 < 2 or 6 < x < 2. h) NT. 2 < x < 6 Ê 1(2) > 1(x) < 1(6) Ê 6 < x < 2 4. NT = necessarily true, NNT = Not necessarily true. Given: 1 < y 5 < 1. a) NT. 1 < y 5 < 1 Ê 1 + 5 < y 5 + 5 < 1 + 5 Ê 4 < y < 6. b) NNT. y = 5 is a counter example. (Actually, never true given that 4 y 6) c) NT. From a), 1 < y 5 < 1, Ê 4 < y < 6 Ê y > 4. d) NT. From a), 1 < y 5 < 1, Ê 4 < y < 6 Ê y < 6. e) NT. 1 < y 5 < 1 Ê 1 + 1 < y 5 + 1 < 1 + 1 Ê 0 < y 4 < 2. f) NT. 1 < y 5 < 1 Ê (1/2)(1 + 5) < (1/2)(y 5 + 5) < (1/2)(1 + 5) Ê 2 < y/2 < 3. g) NT. From a), 4 < y < 6 Ê 1/4 > 1/y > 1/6 Ê 1/6 < 1/y < 1/4. h) NT. 1 < y 5 < 1 Ê y 5 > 1 Ê y > 4 Ê y < 4 Ê y + 5 < 1 Ê (y 5) < 1. Also, 1 < y 5 < 1 Ê y 5 < 1. The pair of inequalities (y 5) < 1 and (y 5) < 1 Ê | y 5 | < 1. 5. 2x 4 Ê x 2 6. 8 3x 5 Ê 3x 3 Ê x Ÿ 1 7. 5x $ Ÿ ( 3x Ê 8x Ÿ 10 Ê x Ÿ
ïïïïïïïïïñqqqqqqqqp x 1 5 4
8. 3(2 x) 2(3 x) Ê 6 3x 6 2x Ê 0 5x Ê 0 x 9. 2x
10.
" #
Ê
" 5
6 x 4
7x
ˆ
10 ‰ 6
3x4 2
7 6
Ê "#
x or
" 3
7 6
ïïïïïïïïïðqqqqqqqqp x 0
5x
x
Ê 12 2x 12x 16
Ê 28 14x Ê 2 x
qqqqqqqqqðïïïïïïïïî x 2
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2 11.
Chapter 1 Preliminaries 4 5
" 3
(x 2)
(x 6) Ê 12(x 2) 5(x 6)
Ê 12x 24 5x 30 Ê 7x 6 or x 67 12. x2 5 Ÿ
123x 4
Ê (4x 20) Ÿ 24 6x
Ê 44 Ÿ 10x Ê 22 5 Ÿ x
qqqqqqqqqñïïïïïïïïî x 22/5
13. y œ 3 or y œ 3 14. y 3 œ 7 or y 3 œ 7 Ê y œ 10 or y œ 4 15. 2t 5 œ 4 or 2t & œ 4 Ê 2t œ 1 or 2t œ 9 Ê t œ "# or t œ 9# 16. 1 t œ 1 or 1 t œ 1 Ê t œ ! or t œ 2 Ê t œ 0 or t œ 2 17. 8 3s œ 18.
s #
9 2
or 8 3s œ #9 Ê 3s œ 7# or 3s œ 25 # Ê sœ
1 œ 1 or
s #
1 œ 1 Ê
s #
œ 2 or
s #
7 6
or s œ
25 6
œ ! Ê s œ 4 or s œ 0
19. 2 x 2; solution interval (2ß 2) 20. 2 Ÿ x Ÿ 2; solution interval [2ß 2]
qqqqñïïïïïïïïñqqqqp x 2 2
21. 3 Ÿ t 1 Ÿ 3 Ê 2 Ÿ t Ÿ 4; solution interval [2ß 4] 22. 1 t 2 1 Ê 3 t 1; solution interval (3ß 1)
qqqqðïïïïïïïïðqqqqp t 3 1
23. % 3y 7 4 Ê 3 3y 11 Ê 1 y solution interval ˆ1ß
11 3
;
11 ‰ 3
24. 1 2y 5 " Ê 6 2y 4 Ê 3 y 2; solution interval (3ß 2) 25. 1 Ÿ
z 5
1Ÿ1 Ê 0Ÿ
z 5
qqqqðïïïïïïïïðqqqqp y 3 2
Ÿ 2 Ê 0 Ÿ z Ÿ 10;
solution interval [0ß 10] 26. 2 Ÿ
1 Ÿ 2 Ê 1 Ÿ solution interval 23 ß 2‘ 3z #
27. "# 3 Ê
2 7
28. 3
" x
x 2 x
2 5
" #
2 7
Ÿ 3 Ê 32 Ÿ z Ÿ 2; qqqqñïïïïïïïïñqqqqp z 2 2/3
Ê 7# x" 5# Ê
7 #
" x
5 #
; solution interval ˆ 27 ß 25 ‰
43 Ê 1
Ê 2x
3z #
Ê
2 7
2 x
( Ê 1
x #
" 7
x 2; solution interval ˆ 27 ß 2‰
qqqqðïïïïïïïïðqqqqp x 2 2/7
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Section 1.1 Real Numbers and the Real Line 29. 2s 4 or 2s 4 Ê s 2 or s Ÿ 2; solution intervals (_ß 2] [2ß _) 30. s 3
" #
or (s 3)
" #
Ê s 5# or s
7 #
Ê s 5# or s Ÿ 7# ; solution intervals ˆ_ß 7# ‘ 5# ß _‰
ïïïïïïñqqqqqqñïïïïïïî s 7/2 5/2
31. 1 x 1 or (" x) 1 Ê x 0 or x 2 Ê x 0 or x 2; solution intervals (_ß !) (2ß _) 32. 2 3x 5 or (2 3x) 5 Ê 3x 3 or 3x 7 Ê x 1 or x 73 ; solution intervals (_ß 1) ˆ 73 ß _‰ 33.
r" #
ïïïïïïðqqqqqqðïïïïïïî x 1 7/3
1 or ˆ r# 1 ‰ 1 Ê r 1 2 or r 1 Ÿ 2
Ê r 1 or r Ÿ 3; solution intervals (_ß 3] [1ß _) 34.
3r 5
"
Ê
or ˆ 3r5 "‰
2 5
or 3r5 53 Ê r 37 or r 1 solution intervals (_ß ") ˆ 73 ß _‰ 3r 5
2 5 7 5
ïïïïïïðqqqqqqðïïïïïïî r 1 7/3
35. x# # Ê kxk È2 Ê È2 x È2 ; solution interval ŠÈ2ß È2‹
qqqqqqðïïïïïïðqqqqqqp x È# È #
36. 4 Ÿ x# Ê 2 Ÿ kxk Ê x 2 or x Ÿ 2; solution interval (_ß 2] [2ß _)
ïïïïïïñqqqqqqñïïïïïïî r 2 2
37. 4 x# 9 Ê 2 kxk 3 Ê 2 x 3 or 2 x 3 Ê 2 x 3 or 3 x 2; solution intervals (3ß 2) (2ß 3) 38.
" 9
x#
Ê
x
" #
" 3
kxk
" #
Ê
" 3
x
or #" x 3" ; solution intervals ˆ "# ß 3" ‰ ˆ 3" ß #" ‰ Ê
" 3
" 4
" #
or
" 3
x
39. (x 1)# 4 Ê kx 1k 2 Ê 2 x 1 2 Ê 1 x 3; solution interval ("ß $)
qqqqðïïïïðqqqqðïïïïðqqqp x 3 2 2 3 " #
qqqqðïïïïðqqqqðïïïïðqqqp x 1/2 1/3 1/3 1/2
qqqqqqðïïïïïïïïðqqqqp x 1 3
40. (x 3)# # Ê kx 3k È2 Ê È2 x 3 È2 or 3 È2 x 3 È2 ; solution interval Š3 È2ß 3 È2‹
qqqqqqðïïïïïïïïðqqqqp x 3 È # 3 È #
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Chapter 1 Preliminaries
41. x# x 0 Ê x# x +
1 4
<
1 4
2 Ê ˆx 12 ‰ <
1 4
ʹx
1 2
¹<
Ê 12 < x
1 2
1 2
<
1 2
Ê 0 < x < 1.
So the solution is the interval (0ß 1) 42. x# x 2 0 Ê x# x +
1 4
9 4
Ê ¹x
1 2
¹
3 2
Ê x
1 2
3 2
or ˆx 12 ‰
3 2
Ê x 2 or x Ÿ 1.
The solution interval is (_ß 1] [2ß _) 43. True if a 0; False if a 0. 44. kx 1k œ 1 x Í k(x 1)k œ 1 x Í 1 x 0 Í x Ÿ 1 45. (1) ka bk œ (a b) or ka bk œ (a b); both squared equal (a b)# (2) ab Ÿ kabk œ kak kbk (3) kak œ a or kak œ a, so kak# œ a# ; likewise, kbk# œ b# (4) x# Ÿ y# implies Èx# Ÿ Èy# or x Ÿ y for all nonnegative real numbers x and y. Let x œ ka bk and y œ kak kbk so that ka bk# Ÿ akak kbkb# Ê ka bk Ÿ kak kbk . 46. If a 0 and b 0, then ab 0 and kabk œ ab œ kak kbk . If a 0 and b 0, then ab 0 and kabk œ ab œ (a)(b) œ kak kbk . If a 0 and b 0, then ab Ÿ 0 and kabk œ (ab) œ (a)(b) œ kak kbk . If a 0 and b 0, then ab Ÿ 0 and kabk œ (ab) œ (a)(b) œ kak kbk . 47. 3 Ÿ x Ÿ 3 and x "# Ê
" #
x Ÿ 3.
48. Graph of kxk kyk Ÿ 1 is the interior of “diamond-shaped" region.
49. Let $ be a real number > 0 and f(x) = 2x + 1. Suppose that | x1 | < $ . Then | x1 | < $ Ê 2| x1 | < 2$ Ê | 2x # | < 2$ Ê | (2x + 1) 3 | < 2$ Ê | f(x) f(1) | < 2$ 50. Let % > 0 be any positive number and f(x) = 2x + 3. Suppose that | x 0 | < % /2. Then 2| x 0 | < % and | 2x + 3 3 | < %. But f(x) = 2x + 3 and f(0) = 3. Thus | f(x) f(0) | < %. 51. Consider: i) a > 0; ii) a < 0; iii) a = 0. i) For a > 0, | a | œ a by definition. Now, a > 0 Ê a < 0. Let a = b. By definition, | b | œ b. Since b = a, | a | œ (a) œ a and | a | œ | a | œ a. ii) For a < 0, | a | œ a. Now, a < 0 Ê a > 0. Let a œ b. By definition, | b | œ b and thus |a| œ a. So again | a | œ |a|. iii) By definition | 0 | œ 0 and since 0 œ 0, | 0 | œ 0. Thus, by i), ii), and iii) | a | œ | a | for any real number. Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 1.2 Lines, Circles and Parabolas Prove | x | > 0 Ê x > a or x < a for any positive number, a. For x 0, | x | œ x. | x | > a Ê x > a. For x < 0, | x | œ x. | x | > a Ê x > a Ê x < a. ii) Prove x > a or x < a Ê | x | > 0 for any positive number, a. a > 0 and x > a Ê | x | œ x. So x > a Ê | x | > a. For a > 0, a < 0 and x < a Ê x < 0 Ê | x | œ x. So x < a Ê x > a Ê | x | > a.
52. i)
53. a)
1=1 Ê |1|=1 ʹb
b)
lal lbl
œ ¹a
†
" b
¹ œ ¹ a¹
† "b ¹ œ
†¹
" b
l bl lbl
¹ œ ¹ a¹
Ê ¹ b¹
† l bl "
† ¹ b" ¹ œ
œ
lbl lbl
Ê
†
¹ b ¹ ¹ "b ¹ ¹ b¹
œ
¹ b¹
†
¹ b¹ ¹ b¹
Ê ¹ b" ¹ œ "
¹ b¹
lal lbl
54. Prove Sn œ kan k œ kakn for any real number a and any positive integer n. ka" k œ kak " œ a, so S" is true. Now, assume that Sk œ ¸ak ¸ œ kak k is true form some positive integer 5 . Since ka" k œ kak " and ¸ak ¸ œ kak k , we have ¸ak" ¸ œ ¸ak † a" ¸ œ ¸ak ¸ka" k œ kak k kak " œ kak k+" . Thus, Sk" œ ¸ak" ¸ œ kak k+" is also true. Thus by the Principle of Mathematical Induction, Sn œ l an l œ l a ln is true for all n positive integers. 1.2 LINES, CIRCLES, AND PARABOLAS 1. ?x œ 1 (3) œ 2, ?y œ 2 2 œ 4; d œ È(?x)# (?y)# œ È4 16 œ 2È5 2. ?x œ $ (1) œ 2, ?y œ 2 (2) œ 4; d œ È(2)# 4# œ 2È5 3. ?x œ 8.1 (3.2) œ 4.9, ?y œ 2 (2) œ 0; d œ È(4.9)# 0# œ 4.9 #
4. ?x œ 0 È2 œ È2, ?y œ 1.5 4 œ 2.5; d œ ÊŠÈ2‹ (2.5)# œ È8.25 5. Circle with center (!ß !) and radius 1.
6. Circle with center (!ß !) and radius È2.
7. Disk (i.e., circle together with its interior points) with center (!ß !) and radius È3. 8. The origin (a single point). 9. m œ
?y ?x
œ
1 2 2 (1)
œ3
perpendicular slope œ "3
10. m œ
?y ?x
œ
# " 2 (2)
œ 34
perpendicular slope œ
4 3
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5
6
Chapter 1 Preliminaries
11. m œ
?y ?x
œ
33 1 2
œ0
12. m œ
14. (a) x œ È2
(b) y œ
# 0 # (#)
; no slope
15. (a) x œ 0
16. (a) x œ 1
(b) y œ È2
(b) y œ 1.3
4 3
œ
perpendicular slope œ 0
perpendicular slope does not exist
13. (a) x œ 1
?y ?x
(b) y œ 0
17. P(1ß 1), m œ 1 Ê y 1 œ 1ax (1)b Ê y œ x 18. P(2ß 3), m œ
" #
Ê y (3) œ
19. P(3ß 4), Q(2ß 5) Ê m œ
?y ?x
20. P(8ß 0), Q(1ß 3) Ê m œ
œ
?y ?x
" #
(x 2) Ê y œ
54 2 3
œ
" #
x4
œ "5 Ê y 4 œ "5 (x 3) Ê y œ "5 x
30 1 (8)
œ
3 7
Ê y0œ
3 7
ax (8)b Ê y œ
3 7
23 5
x
21. m œ 54 , b œ 6 Ê y œ 54 x 6
22. m œ "# , b œ 3 Ê y œ
" #
23. m œ 0, P(12ß 9) Ê y œ 9
24. No slope, P ˆ "3 ß %‰ Ê x œ
24 7
x3 " 3
25. a œ 1, b œ 4 Ê (0ß 4) and ("ß 0) are on the line Ê m œ
?y ?x
œ
04 1 0
œ 4 Ê y œ 4x 4
26. a œ 2, b œ 6 Ê (2ß 0) and (!ß 6) are on the line Ê m œ
?y ?x
œ
6 0 02
œ 3 Ê y œ 3x 6
27. P(5ß 1), L: 2x 5y œ 15 Ê mL œ 25 Ê parallel line is y (1) œ 25 (x 5) Ê y œ 25 x 1 È È È 28. P ŠÈ2ß 2‹ , L: È2x 5y œ È3 Ê mL œ 52 Ê parallel line is y 2 œ 52 Šx ŠÈ2‹‹ Ê y œ 52 x
8 5
29. P(4ß 10), L: 6x 3y œ 5 Ê mL œ 2 Ê m¼ œ "# Ê perpendicular line is y 10 œ "# (x 4) Ê y œ "# x 12 30. P(!ß 1), L: 8x 13y œ 13 Ê mL œ
8 13
13 Ê m¼ œ 13 8 Ê perpendicular line is y œ 8 x 1
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Section 1.2 Lines, Circles and Parabolas 31. x-intercept œ 4, y-intercept œ 3
32. x-intercept œ 4, y-intercept œ 2
33. x-intercept œ È3, y-intercept œ È2
34. x-intercept œ 2, y-intercept œ 3
35. Ax By œ C" Í y œ AB x
C" B
and Bx Ay œ C# Í y œ
B A
x
C# A.
Since ˆ AB ‰ ˆ AB ‰ œ 1 is the
product of the slopes, the lines are perpendicular. 36. Ax By œ C" Í y œ AB x slope
AB ,
C" B
and Ax By œ C# Í y œ AB x
C# B.
Since the lines have the same
they are parallel.
37. New position œ axold ?xß yold ?yb œ (# &ß 3 (6)) œ ($ß 3). 38. New position œ axold ?xß yold ?yb œ (6 (6)ß 0 0) œ (0ß 0). 39. ?x œ 5, ?y œ 6, B(3ß 3). Let A œ (xß y). Then ?x œ x# x" Ê 5 œ 3 x Ê x œ 2 and ?y œ y# y" Ê 6 œ 3 y Ê y œ 9. Therefore, A œ (#ß 9). 40. ?x œ " " œ !, ?y œ ! ! œ !
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8
Chapter 1 Preliminaries
41. C(!ß 2), a œ 2 Ê x# (y 2)# œ 4
42. C($ß 0), a œ 3 Ê (x 3)# y# œ 9
43. C(1ß 5), a œ È10 Ê (x 1)# (y 5)# œ 10
44. C("ß "), a œ È2 Ê (x 1)# (y 1)# œ 2 x œ 0 Ê (0 1)# (y 1)# œ 2 Ê (y 1)# œ 1 Ê y 1 œ „ 1 Ê y œ 0 or y œ 2. Similarly, y œ 0 Ê x œ 0 or x œ 2
#
45. C ŠÈ3ß 2‹ , a œ 2 Ê Šx È3‹ (y 2)# œ 4, #
x œ 0 Ê Š0 È3‹ (y 2)# œ 4 Ê (y 2)# œ 1 Ê y 2 œ „ 1 Ê y œ 1 or y œ 3. Also, y œ 0 #
#
Ê Šx È3‹ (0 2)# œ 4 Ê Šx È3‹ œ 0 Ê x œ È 3 # 46. C ˆ3ß "# ‰, a œ 5 Ê (x 3)# ˆy "# ‰ œ 25, so # x œ 0 Ê (0 3)# ˆy "# ‰ œ 25 # Ê ˆy "# ‰ œ 16 Ê y
" #
œ „4 Ê yœ
9 #
# or y œ 7# . Also, y œ 0 Ê (x 3)# ˆ0 "# ‰ œ 25
Ê (x 3)# œ Ê xœ3„
99 4 3È11 #
Ê x3œ „
3È11 #
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Section 1.2 Lines, Circles and Parabolas 47. x# y# 4x 4y % œ 0 Ê x# %B y# 4y œ 4 Ê x# 4x 4 y# 4y 4 œ 4 Ê (x 2)# (y 2)# œ 4 Ê C œ (2ß 2), a œ 2.
48. x# y# 8x 4y 16 œ 0 Ê x# 8x y# 4y œ 16 Ê x# 8x 16 y# 4y 4 œ 4 Ê (x 4)# (y 2)# œ 4 Ê C œ (%ß 2), a œ 2.
49. x# y# 3y 4 œ 0 Ê x# y# 3y œ 4 Ê x# y# 3y 94 œ 25 4 # Ê x# ˆy 3# ‰ œ
25 4
Ê C œ ˆ0ß 3# ‰ ,
a œ 5# .
50. x# y# 4x #
9 4 #
œ0
Ê x 4x y œ
9 4 #
Ê x# 4x 4 y œ Ê (x 2)# y# œ
25 4
25 4
Ê C œ (2ß 0), a œ 5# .
51. x# y# 4x 4y œ 0 Ê x# 4x y# 4y œ 0 Ê x# 4x 4 y# 4y 4 œ 8 Ê (x 2)# (y 2)# œ 8 Ê C(2ß 2), a œ È8.
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9
10
Chapter 1 Preliminaries
52. x# y# 2x œ 3 Ê x# 2x 1 y# œ 4 Ê (x 1)# y# œ 4 Ê C œ (1ß 0), a œ 2.
2 53. x œ #ba œ 2(1) œ1
Ê y œ (1)# 2(1) 3 œ 4 Ê V œ ("ß 4). If x œ 0 then y œ 3. Also, y œ 0 Ê x# 2x 3 œ 0 Ê (x 3)(x 1) œ 0 Ê x œ 3 or x œ 1. Axis of parabola is x œ 1.
4 54. x œ #ba œ 2(1) œ 2
Ê y œ (2)# 4(2) 3 œ 1 Ê V œ (2ß 1). If x œ 0 then y œ 3. Also, y œ 0 Ê x# 4x 3 œ 0 Ê (x 1)(x 3) œ 0 Ê x œ 1 or x œ 3. Axis of parabola is x œ 2.
55. x œ #ba œ 2(4 1) œ 2 Ê y œ (2)# 4(2) œ 4 Ê V œ (2ß 4). If x œ 0 then y œ 0. Also, y œ 0 Ê x# 4x œ 0 Ê x(x 4) œ 0 Ê x œ 4 or x œ 0. Axis of parabola is x œ 2.
56. x œ #ba œ 2(4 1) œ 2 Ê y œ (2)# 4(2) 5 œ 1 Ê V œ (2ß 1). If x œ 0 then y œ 5. Also, y œ 0 Ê x# 4x 5 œ 0 Ê x# 4x 5 œ 0 Ê x œ
4 „È 4 #
Ê no x intercepts. Axis of parabola is x œ 2.
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Section 1.2 Lines, Circles and Parabolas 57. x œ #ba œ 2(61) œ 3 Ê y œ (3)# 6(3) 5 œ 4 Ê V œ (3ß %). If x œ 0 then y œ 5. Also, y œ 0 Ê x# 6x 5 œ 0 Ê (x 5)(x 1) œ 0 Ê x œ 5 or x œ 1. Axis of parabola is x œ 3.
1 58. x œ #ba œ 2(2) œ
" 4
#
Ê y œ 2 ˆ "4 ‰ 4" 3 œ 23 8 ‰ Ê V œ ˆ "4 ß 23 . If x œ 0 then y œ 3. 8
Also, y œ 0 Ê 2x# x 3 œ 0 Ê xœ
1„È23 4
Ê no x intercepts.
Axis of parabola is x œ "4 .
1 59. x œ #ba œ 2(1/2) œ 1 " #
(1)# (1) 4 œ 72 Ê V œ ˆ"ß 72 ‰ . If x œ 0 then y œ 4. Ê yœ
Also, y œ 0 Ê Ê xœ
1 „ È 7 1
" #
x# x 4 œ 0 Ê no x intercepts.
Axis of parabola is x œ 1. 60. x œ #ba œ 2(21/4) œ 4 Ê y œ "4 (4)# 2(4) 4 œ 8 Ê V œ (4ß 8) . If x œ 0 then y œ 4. Also, y œ 0 Ê "4 x# 2x 4 œ 0 Ê xœ
2 „ È 8 1/2
œ 4 „ 4È2.
Axis of parabola is x œ 4.
61. The points that lie outside the circle with center (!ß 0) and radius È7. 62. The points that lie inside the circle with center (!ß 0) and radius È5. 63. The points that lie on or inside the circle with center ("ß 0) and radius 2. 64. The points lying on or outside the circle with center (!ß 2) and radius 2. 65. The points lying outside the circle with center (!ß 0) and radius 1, but inside the circle with center (!ß 0), and radius 2 (i.e., a washer).
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11
12
Chapter 1 Preliminaries
66. The points on or inside the circle centered at (!ß !) with radius 2 and on or inside the circle centered at (2ß 0) with radius 2.
67. x# y# 6y 0 Ê x# (y 3)# 9. The interior points of the circle centered at (!ß 3) with radius 3, but above the line y œ 3.
68. x# y# 4x 2y 4 Ê (x 2)# (y 1)# 9. The points exterior to the circle centered at (2ß 1) with radius 3 and to the right of the line x œ 2.
69. (x 2)# (y 1)# 6
70. (x 4)# (y 2)# 16
71. x# y# Ÿ 2, x 1
72. x# y# 4, (x 1)# (y 3)# 10
73. x# y# œ 1 and y œ 2x Ê 1 œ x# 4x# œ 5x# Ê Šx œ
" È5
and y œ
2 È5 ‹
or Šx œ È"5 and y œ È25 ‹ .
Thus, A Š È"5 ß È25 ‹ , B Š È"5 ß È25 ‹ are the points of intersection.
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Section 1.2 Lines, Circles, and Parabolas 74. x y œ 1 and (x 1)# y# œ 1 Ê 1 œ (y)# y# œ 2y# Ê Šy œ
" È2
and x œ "
Šy œ È"2 and x œ 1 A Š"
" È2
" È2 ‹
" È2 ‹ .
ß È"2 ‹ and B Š1
or
Thus,
" È2
ß È"2 ‹
are intersection points.
75. y x œ 1 and y œ x# Ê x# x œ 1 1 „È 5 . # 1 È 5 3 È 5 If x œ # , then y œ x 1 œ # . È È If x œ 1# 5 , then y œ x 1 œ 3# 5 . È È È È Thus, A Š 1# 5 ß 3# 5 ‹ and B Š 1# 5 ß 3# 5 ‹
Ê x# x 1 œ 0 Ê x œ
are the intersection points.
76. y œ x and C œ (x 1)# Ê (x 1)# œ x 3 „È 5 . # È 5 3 3 È 5 x œ # , then y œ x œ # . If È È x œ 3# 5 , then y œ x œ 3# 5 . È È È Thus, A Š 3# 5 ß 5#3 ‹ and B Š 3# 5
Ê x# 3x " œ 0 Ê x œ
If
È
ß 3# 5 ‹
are the intersection points.
77. y œ 2x# 1 œ x# Ê 3x# œ 1 Ê x œ È"3 and y œ 3" or x œ È"3 and y œ 3" . Thus, A Š È"3 ß 3" ‹ and B Š È"3 ß 3" ‹ are the intersection points.
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13
14
Chapter 1 Preliminaries
78. y œ
x# 4
œ (x 1)# Ê 0 œ #
3x# 4
2x 1
Ê 0 œ 3x 8x 4 œ (3x 2)(x 2) Ê x œ 2 and y œ yœ
#
x 4
x# 4
œ 1, or x œ
œ 9" . Thus, A(2ß 1) and
2 3 and 2 B ˆ 3 ß 9" ‰
are the intersection points.
79. x# y# œ 1 œ (x 1)# y# Ê x# œ (x 1)# œ x# 2x 1 Ê 0 œ 2x 1 Ê x œ "# . Hence y# œ " x # œ A Š "# ß
È3 # ‹
and
3 4
or y œ „
È3 #
È B Š "# ß #3 ‹
. Thus,
are the
intersection points.
80. x# y# œ 1 œ x# y Ê y# œ y Ê y(y 1) œ 0 Ê y œ 0 or y œ 1. If y œ 1, then x# œ " y# œ 0 or x œ 0. If y œ 0, then x# œ 1 y# œ 1 or x œ „ 1. Thus, A(0ß 1), B("ß 0), and C(1ß 0) are the intersection points.
81. (a) A ¸ (69°ß 0 in), B ¸ (68°ß .4 in) Ê m œ (b) A ¸ (68°ß .4 in), B ¸ (10°ß 4 in) Ê m œ (c) A ¸ (10°ß 4 in), B ¸ (5°ß 4.6 in) Ê m œ 82. The time rate of heat transfer across a material, to the temperature gradient across the material, of the material.
?U ?>
œ
X -kA ? ?B
Ê
?U ÎA k = ??> X . ?B
68° 69° .4 0 ¸ 2.5°/in. 10° 68° 4 .4 ¸ 16.1°/in. 5° 10° 4.6 4 ¸ 8.3°/in. ?U ?> , is directly ?X ?B (the slopes
Note that
?U ?>
proportional to the cross-sectional area, A, of the material, from the previous problem), and to a constant characteristic
and
?X ?B
are of opposite sign because heat flow is toward lower
temperature. So a small value of k corresponds to low heat flow through the material and thus the material is a good insulator.Since all three materials have the same cross section and the heat flow across each is the same (temperatures are X not changing), we may define another constant, K, characteristics of the material: K œ ?"X Þ Using the values of ? ?B from ?B
the prevous problem, fiberglass has the smallest K at 0.06 and thus is the best insulator. Likewise, the wallboard is the poorest insulator, with K œ 0.4. 83. p œ kd 1 and p œ 10.94 at d œ 100 Ê k œ
10.94" 100
œ 0.0994. Then p œ 0.0994d 1 is the diver's
pressure equation so that d œ 50 Ê p œ (0.0994)(50) 1 œ 5.97 atmospheres. 84. The line of incidence passes through (!ß 1) and ("ß 0) Ê The line of reflection passes through ("ß 0) and (#ß ") 0 Ê m œ 1# 1 œ 1 Ê y 0 œ 1(x 1) Ê y œ x 1 is the line of reflection.
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Section 1.2 Lines, Circles, and Parabolas 85. C œ
5 9
(F 32) and C œ F Ê F œ
86. m œ
37.1 100
œ
14 ?x
Ê ?x œ
14 .371 .
5 9
F
160 9
Ê
4 9
15
F œ 160 9 or F œ 40° gives the same numerical reading.
#
14 ‰ Therefore, distance between first and last rows is É(14)# ˆ .371 ¸ 40.25 ft.
87. length AB œ È(5 1)# (5 2)# œ È16 9 œ 5 length AC œ È(4 1)# (# #)# œ È9 16 œ 5 length BC œ È(4 5)# (# 5)# œ È1 49 œ È50 œ 5È2 Á 5 #
88. length AB œ Ê(1 0)# ŠÈ3 0‹ œ È1 3 œ 2 length AC œ È(2 0)# (0 0)# œ È4 0 œ 2 #
length BC œ Ê(2 1)# Š0 È3‹ œ È1 3 œ 2 89. Length AB œ È(?x)# (?y)# œ È1# 4# œ È17 and length BC œ È(?x)# (?y)# œ È4# 1# œ È17. Also, slope AB œ 41 and slope BC œ "4 , so AB ¼ BC. Thus, the points are vertices of a square. The coordinate increments from the fourth vertex D(xß y) to A must equal the increments from C to B Ê 2 x œ ?x œ 4 and 1 y œ ?y œ " Ê x œ 2 and y œ 2. Thus D(#ß 2) is the fourth vertex.
90. Let A œ (xß 2) and C œ (9ß y) Ê B œ (xß y). Then 9 x œ kADk and 2 y œ kDCk Ê 2(9 x) 2(2 y) œ 56 and 9 x œ 3(2 y) Ê 2(3(2 y)) 2(2 y) œ 56 Ê y œ 5 Ê 9 x œ 3(2 (5)) Ê x œ 12. Therefore, A œ (12ß 2), C œ (9ß 5), and B œ (12ß 5). 91. Let A("ß "), B(#ß $), and C(2ß !) denote the points. Since BC is vertical and has length kBCk œ 3, let D" ("ß 4) be located vertically upward from A and D# ("ß 2) be located vertically downward from A so that kBCk œ kAD" k œ kAD# k œ 3. Denote the point D$ (xß y). Since the slope of AB equals the slope of 3 " CD$ we have yx 2 œ 3 Ê 3y 9 œ x 2 or
x 3y œ 11. Likewise, the slope of AC equals the slope 0 2 of BD$ so that yx 2 œ 3 Ê 3y œ 2x 4 or 2x 3y œ 4.
Solving the system of equations
x 3y œ "" we find x œ 5 and y œ 2 yielding the vertex D$ (5ß #). 2x 3y œ 4
92. Let ax, yb, x Á ! and/or y Á ! be a point on the coordinate plane. The slope, m, of the segment a!ß !b to ax, yb is yx . A 90‰
rotation gives a segment with slope mw œ m" œ xy . If this segment has length equal to the original segment, its endpoint will be ay, xb or ay, xb, the first of these corresponds to a counter-clockwise rotation, the latter to a clockwise rotation. (a) ("ß 4); (b) (3ß 2); (c) (5ß 2); (d) (0ß x);
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16
Chapter 1 Preliminaries (e) (yß 0);
(f) (yß x);
(g) (3ß 10)
93. 2x ky œ 3 has slope 2k and 4x y œ 1 has slope 4. The lines are perpendicular when 2k (4) œ 1 or k œ 8 and parallel when 2k œ 4 or k œ "# .
94. At the point of intersection, 2x 4y œ 6 and 2x 3y œ 1. Subtracting these equations we find 7y œ 7 or y œ 1. Substitution into either equation gives x œ 1 Ê (1ß 1) is the intersection point. The line through (1ß 1) and ("ß #) is vertical with equation x œ 1. 95. Let M(aß b) be the midpoint. Since the two triangles shown in the figure are congruent, the value a must lie midway between x" and x# , so a œ x" #x# . Similarly, b œ
y " y # # .
96. (a) L has slope 1 so M is the line through P(2ß 1) with slope 1; or the line y œ x 3. At the intersection point, Q, we have equal y-values, y œ x 2 œ x 3. Thus, 2x œ 1 or x œ "# . Hence Q has coordinates ˆ "# ß 5# ‰ . The distance from P to L œ the distance from P to Q œ Ɉ #3 ‰# ˆ 3# ‰# œ É 18 4 œ (b) L has slope 43 so M has slope
3 4
3È 2 # .
and M has the equation 4y 3x œ 12. We can rewrite the equations of
84 the lines as L: x y œ 3 and M: B 43 y œ 4. Adding these we get 25 12 y œ 7 so y œ 25 . Substitution 12 ‰ ˆ 12 84 ‰ into either equation gives x œ 43 ˆ 84 25 4 œ 25 so that Q 25 ß 25 is the point of intersection. The distance 3 4
from P to L œ Ɉ4
12 ‰# 25
ˆ6
84 ‰# 25
œ
22 5 .
(c) M is a horizontal line with equation y œ b. The intersection point of L and M is Q("ß b). Thus, the distance from P to L is È(a 1)# 0# œ ka 1k . (d) If B œ 0 and A Á 0, then the distance from P to L is ¸ AC x! ¸ as in (c). Similarly, if A œ 0 and B Á 0, the distance is ¸ CB y! ¸ . If both A and B are Á 0 then L has slope AB so M has slope AB . Thus, L: Ax By œ C and M: Bx Ay œ Bx! Ay! . Solving these equations simultaneously we find the point of intersection Q(xß y) with x œ
ACB aAy! Bx! b A# B#
P to Q equals È(?x)# (?y)# , where (?x)# œ œ
A# aAx! By! Cb# aA# B# b#
#
#
BCA aAy! Bx! b . A# B# # # # # ABy! B x! Š x! aA B bAAC ‹ # B#
and y œ
#
#
A y! ABx! , and (?y)# œ Š y! aA B bABC ‹ œ # B# #
! Cb Thus, È(?x)# (?y)# œ É aAx!A#By œ B#
kAx! By! Ck ÈA# B#
The distance from
B# aAx! By! Cb# . aA# B# b#
.
1.3 FUNCTIONS AND THEIR GRAPHS 1. domain œ (_ß _); range œ [1ß _) 3. domain œ (!ß _); y in range Ê y œ
2. domain œ [0ß _); range œ (_ß 1] " Èt
, t 0 Ê y# œ
" t
and y ! Ê y can be any positive real number
Ê range œ (!ß _).
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Section 1.3 Functions and Their Graphs 4. domain œ [0ß _); y in range Ê y œ
" 1 È t
17
, t 0. If t œ 0, then y œ 1 and as t increases, y becomes a smaller
and smaller positive real number Ê range œ (0ß 1]. 5. 4 z# œ (2 z)(2 z) 0 Í z − [2ß 2] œ domain. Largest value is g(0) œ È4 œ 2 and smallest value is g(2) œ g(2) œ È0 œ 0 Ê range œ [0ß 2]. 6. domain œ (2ß 2) from Exercise 5; smallest value is g(0) œ "# and as 0 z increases to 2, g(z) gets larger and larger (also true as z 0 decreases to 2) Ê range œ "# ß _‰ . 7. (a) Not the graph of a function of x since it fails the vertical line test. (b) Is the graph of a function of x since any vertical line intersects the graph at most once. 8. (a) Not the graph of a function of x since it fails the vertical line test. (b) Not the graph of a function of x since it fails the vertical line test. 9. y œ Ɉ "x ‰ " Ê (a) No (x !Ñ; (c) No; if x ",
" x
" x
" ! Ê x Ÿ 1 and x !. So,
"Ê
" x
(b) No; division by ! undefined; (d) Ð!ß "Ó
" !;
10. y œ É# Èx Ê # Èx ! Ê Èx ! and Èx Ÿ #. Èx ! Ê x ! and Èx Ÿ # Ê x Ÿ %Þ So, ! Ÿ x Ÿ %. (a) No; (b) No; (c) Ò!ß %Ó #
11. base œ x; (height)# ˆ #x ‰ œ x# Ê height œ
È3 #
x; area is a(x) œ
" #
(base)(height) œ
" #
(x) Š
È3 # x‹
œ
È3 4
x# ;
perimeter is p(x) œ x x x œ 3x. 12. s œ side length Ê s# s# œ d# Ê s œ
d È2
; and area is a œ s# Ê a œ
" #
d#
13. Let D œ diagonal of a face of the cube and j œ the length of an edge. Then j# D# œ d# and (by Exercise 10) D# œ 2j# Ê 3j# œ d# Ê j œ
d È3
. The surface area is 6j# œ
6d# 3
14. The coordinates of P are ˆxß Èx‰ so the slope of the line joining P to the origin is m œ ˆx, Èx‰ œ ˆ m"# ,
#
œ 2d# and the volume is j$ œ Š d3 ‹ Èx x
œ
" Èx
"‰ m .
15. The domain is a_ß _b.
16. The domain is a_ß _b.
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$Î#
œ
d$ 3È 3
(x 0). Thus,
.
18
Chapter 1 Preliminaries
17. The domain is a_ß _b.
18. The domain is Ð_ß !Ó.
19. The domain is a_ß !b a!ß _b.
20. The domain is a_ß !b a!ß _b.
21. Neither graph passes the vertical line test (a)
(b)
22. Neither graph passes the vertical line test (a)
(b)
Ú xyœ" Þ Ú yœ1x Þ or or kx yk œ 1 Í Û Í Û ß ß Ü x y œ " à Ü y œ " x à
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 1.3 Functions and Their Graphs 23.
x y
0 0
25. y œ œ
1 1
2 0
24.
x y
0 1
1 0
2 0
" , x0 26. y œ œ x x, 0 Ÿ x
3 x, x Ÿ 1 2x, 1 x
27. (a) Line through a!ß !b and a"ß "b: y œ x Line through a"ß "b and a#ß !b: y œ x 2 x, 0 Ÿ x Ÿ 1 f(x) œ œ x 2, 1 x Ÿ 2 Ú Ý Ý 2, ! Ÿ x " !ß " Ÿ x # (b) f(x) œ Û Ý Ý 2ß # Ÿ x $ Ü !ß $ Ÿ x Ÿ % 28. (a) Line through a!ß 2b and a#ß !b: y œ x 2 " Line through a2ß "b and a&ß !b: m œ !& # œ x #, 0 x Ÿ # f(x) œ œ " $ x &$ , # x Ÿ &
" $
$ ! ! Ð"Ñ œ " $ % #! œ #
(b) Line through a"ß !b and a!ß $b: m œ Line through a!ß $b and a#ß "b: m œ f(x) œ œ
œ "$ , so y œ "$ ax 2b " œ "$ x
& $
$, so y œ $x $ œ #, so y œ #x $
$x $, " x Ÿ ! #x $, ! x Ÿ #
29. (a) Line through a"ß "b and a!ß !b: y œ x Line through a!ß "b and a"ß "b: y œ " Line through a"ß "b and a$ß !b: m œ !" $" œ Ú x " Ÿ x ! " !xŸ" f(x) œ Û Ü "# x $# "x$
" #
œ "# , so y œ "# ax "b " œ "# x
$ #
(b) Line through a#ß "b and a!ß !b: y œ "# x
Line through a!ß #b and a"ß !b: y œ #x # Line through a"ß "b and a$ß "b: y œ "
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19
20
Chapter 1 Preliminaries Ú
" #x
f(x) œ Û #x # Ü "
# Ÿ x Ÿ ! !xŸ" "xŸ$
30. (a) Line through ˆ T# ß !‰ and aTß "b: m œ
Ú A, Ý Ý Ý Aß f(x) œ Û Aß Ý Ý Ý Ü Aß
! Ÿ x T# T # Ÿ x T T Ÿ x $#T $T # Ÿ x Ÿ #T x #
31. (a) From the graph, (b)
x #
1
x 0:
x #
x 0:
x 2
œ T# , so y œ T# ˆx T# ‰ 0 œ T# x "
!, 0 Ÿ x Ÿ T# # T T x ", # x Ÿ T
f(x) œ
(b)
"! TaTÎ#b
4 x
1
x #
Ê 4 x
1
4 x
Ê x − (2ß 0) (%ß _)
1 4x 0 # 2x8 0 Ê x 2x
0 Ê
(x4)(x2) #x
0
(x4)(x2) #x
0
Ê x 4 since x is positive; 1
4 x
0 Ê
x# 2x8 2x
0 Ê
Ê x 2 since x is negative; sign of (x 4)(x 2) ïïïïïðïïïïïðïïïïî 2 % Solution interval: (#ß 0) (%ß _)
3 2 x 1 x 1 3 2 x 1 x 1
32. (a) From the graph, (b) Case x 1:
Ê x − (_ß 5) (1ß 1) Ê
3(x1) x 1
2
Ê 3x 3 2x 2 Ê x 5. Thus, x − (_ß 5) solves the inequality. Case 1 x 1:
3 x 1
2 x 1
Ê
3(x1) x 1
2
Ê 3x 3 2x 2 Ê x 5 which is true if x 1. Thus, x − (1ß 1) solves the inequality. Case 1 x: x3 1 x2 1 Ê 3x 3 2x 2 Ê x 5 which is never true if 1 x, so no solution here. In conclusion, x − (_ß 5) (1ß 1). 33. (a) ÚxÛ œ 0 for x − [0ß 1)
(b) ÜxÝ œ 0 for x − (1ß 0]
34. ÚxÛ œ ÜxÝ only when x is an integer. 35. For any real number x, n Ÿ x Ÿ n ", where n is an integer. Now: n Ÿ x Ÿ n " Ê Ðn "Ñ Ÿ x Ÿ n. By definition: ÜxÝ œ n and ÚxÛ œ n Ê ÚxÛ œ n. So ÜxÝ œ ÚxÛ for all x − d .
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 1.3 Functions and Their Graphs
21
36. To find f(x) you delete the decimal or fractional portion of x, leaving only the integer part.
37. v œ f(x) œ xÐ"% 2xÑÐ22 2xÑ œ %x$ 72x# $!)x; ! x 7Þ 38. (a) Let h œ height of the triangle. Since the triangle is isosceles, AB # AB # œ 2# Ê AB œ È2Þ So, #
h# "# œ ŠÈ2‹ Ê h œ " Ê B is at a!ß "b Ê slope of AB œ " Ê The equation of AB is y œ f(x) œ B "; x − Ò!ß "Ó. (b) AÐxÑ œ 2x y œ 2xÐx "Ñ œ 2x# #x; x − Ò!ß "Ó. 39. (a) Because the circumference of the original circle was )1 and a piece of length x was removed. x x (b) r œ )1# 1 œ % #1 (c) h œ È"' r# œ É"' ˆ% #
x‰ (d) V œ "$ 1 r# h œ "$ 1ˆ )1# † 1
x ‰# #1
œ É"' ˆ16
È"'1x x# #1
œ
4x 1
x# ‰ %1#
œ É 4x 1
x# %1#
œ É "'%11#x
x# %1#
œ
È"'1xx# #1
a)1 xb# È"'1x x# #%1#
40. (a) Note that 2 mi = 10,560 ft, so there are È)!!# x# feet of river cable at $180 per foot and a"!ß &'! xb feet of land cable at $100 per foot. The cost is Caxb œ ")!È)!!# x# "!!a"!ß &'! xb. (b) Ca!b œ $"ß #!!ß !!! Ca&!!b ¸ $"ß "(&ß )"# Ca"!!!b ¸ $"ß ")'ß &"# Ca"&!!b ¸ $"ß #"#ß !!! Ca#!!!b ¸ $"ß #%$ß ($# Ca#&!!b ¸ $"ß #()ß %(* Ca$!!!b ¸ $"ß $"%ß )(! Values beyond this are all larger. It would appear that the least expensive location is less than 2000 feet from the point P. 41. A curve symmetric about the x-axis will not pass the vertical line test because the points ax, yb and ax, yb lie on the same vertical line. The graph of the function y œ faxb œ ! is the x-axis, a horizontal line for which there is a single y-value, !, for any x. 42. Pick 11, for example: "" & œ "' Ä # † "' œ $# Ä $# ' œ #' Ä faxb œ
#ax&b' #
#' #
œ "$ Ä "$ # œ "", the original number.
# œ x, the number you started with.
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22
Chapter 1 Preliminaries
1.4 IDENTIFYING FUNCTIONS; MATHEMATICAL MODELS 1. (a) linear, polynomial of degree 1, algebraic. (c) rational, algebraic.
(b) power, algebraic. (d) exponential.
2. (a) polynomial of degree 4, algebraic. (c) algebraic.
(b) exponential. (d) power, algebraic.
3. (a) rational, algebraic. (c) trigonometric.
(b) algebraic. (d) logarithmic.
4. (a) logarithmic. (c) exponential.
(b) algebraic. (d) trigonometric.
5. (a) Graph h because it is an even function and rises less rapidly than does Graph g. (b) Graph f because it is an odd function. (c) Graph g because it is an even function and rises more rapidly than does Graph h. 6. (a) Graph f because it is linear. (b) Graph g because it contains a!ß "b. (c) Graph h because it is a nonlinear odd function. 7. Symmetric about the origin Dec: _ x _ Inc: nowhere
8. Symmetric about the y-axis Dec: _ x ! Inc: ! x _
9. Symmetric about the origin Dec: nowhere Inc: _ x ! !x_
10. Symmetric about the y-axis Dec: ! x _ Inc: _ x !
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Section 1.4 Identifying Functions; Mathematical Models 11. Symmetric about the y-axis Dec: _ x Ÿ ! Inc: ! x _
12. No symmetry Dec: _ x Ÿ ! Inc: nowhere
13. Symmetric about the origin Dec: nowhere Inc: _ x _
14. No symmetry Dec: ! Ÿ x _ Inc: nowhere
15. No symmetry Dec: ! Ÿ x _ Inc: nowhere
16. No symmetry Dec: _ x Ÿ ! Inc: nowhere
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23
24
Chapter 1 Preliminaries
17. Symmetric about the y-axis Dec: _ x Ÿ ! Inc: ! x _
18. Symmetric about the y-axis Dec: ! Ÿ x _ Inc: _ x !
19. Since a horizontal line not through the origin is symmetric with respect to the y-axis, but not with respect to the origin, the function is even. 20. faxb œ x& œ
" x&
and faxb œ axb& œ
" a x b&
œ ˆ x"& ‰ œ faxb. Thus the function is odd.
21. Since faxb œ x# " œ axb# " œ faxb. The function is even. 22. Since Òfaxb œ x# xÓ Á Òfaxb œ axb# xÓ and Òfaxb œ x# xÓ Á Òfaxb œ axb# xÓ the function is neither even nor odd. 23. Since gaxb œ x$ x, gaxb œ x$ x œ ax$ xb œ gaxb. So the function is odd. 24. gaxb œ x% $x# " œ axb% $axb# " œ gaxbß thus the function is even. 25. gaxb œ
" x# "
26. gaxb œ
x x# " ;
27. hatb œ
" t ";
œ
" axb# "
œ gaxb. Thus the function is even.
gaxb œ x#x" œ gaxb. So the function is odd.
h a t b œ
" t " ;
h at b œ
" " t.
Since hatb Á hatb and hatb Á hatb, the function is neither even nor odd.
28. Since l t$ | œ l atb$ |, hatb œ hatb and the function is even. 29. hatb œ 2t ", hatb œ 2t ". So hatb Á hatb. hatb œ 2t ", so hatb Á hatb. The function is neither even nor odd. 30. hatb œ 2l t l " and hatb œ 2l t l " œ 2l t l ". So hatb œ hatb and the function is even. 31. (a)
The graph supports the assumption that y is proportional to x. The constant of proportionality is estimated from the slope of the regression line, which is 0.166.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 1.4 Identifying Functions; Mathematical Models (b)
25
The graph supports the assumption that y is proportional to x"Î# . The constant of proportionality is estimated from the slope of the regression line, which is 2.03.
32. (a) Because of the wide range of values of the data, two graphs are needed to observe all of the points in relation to the regression line.
The graphs support the assumption that y is proportional to $x . The constant of proportionality is estimated from the slope of the regression line, which is 5.00. (b) The graph supports the assumption that y is proportional to ln x. The constant of proportionality is extimated from the slope of the regression line, which is 2.99.
33. (a) The scatterplot of y œ reaction distance versus x œ speed is
Answers for the constant of proportionality may vary. The constant of proportionality is the slope of the line, which is approximately 1.1.
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26
Chapter 1 Preliminaries (b) Calculate x w œ speed squared. The scatterplot of x w versus y œ braking distance is:
Answers for the constant of proportionality may vary. The constant of proportionality is the slope of the line, which is approximately 0.059. 34. Kepler's 3rd Law is Tadaysb œ !Þ%"R$Î# , R in millions of miles. "Quaoar" is 4 ‚ "!* miles from Earth, or about 4 ‚ "!* *$ ‚ "!' ¸ % ‚ "!* miles from the sun. Let R œ 4000 (millions of miles) and T œ a!Þ%"ba%!!!b$Î# days ¸ "!$ß (#$ days. 35. (a)
The hypothesis is reasonable. (b) The constant of proportionality is the slope of the line ¸ (c) y(in.) œ a!Þ)( in./unit massba"$ unit massb œ ""Þ$" in. 36. (a)
)Þ(%" ! "! !
in./unit mass œ !Þ)(% in./unit mass.
(b)
Graph (b) suggests that y œ k x$ is the better model. This graph is more linear than is graph (a). 1.5 COMBINING FUNCTIONS; SHIFTING AND SCALING GRAPHS 1. Df : _ x _, Dg : x 1 Ê Dfbg œ Dfg : x 1. Rf : _ y _, Rg : y 0, Rfbg : y 1, Rfg : y 0 2. Df : x 1 0 Ê x 1, Dg : x 1 0 Ê x 1. Therefore Dfbg œ Dfg : x 1. Rf œ Rg : y 0, Rfbg : y È2, Rfg : y 0
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 1.5 Combining Functions; Shifting and Scaling Graphs 3. Df : _ x _, Dg : _ x _ Ê DfÎg : _ x _ since g(x) Á 0 for any x; DgÎf : _ x _ since f(x) Á 0 for any x. Rf : y œ 2, Rg : y 1, RfÎg : 0 y Ÿ 2, RgÎf : y "# 4. Df : _ x _, Dg : x 0 Ê DfÎg : x 0 since g(x) Á 0 for any x 0; DgÎf : x 0 since f(x) Á 0 for any x 0. Rf : y œ 1, Rg : y 1, RfÎg : 0 y Ÿ 1, RgÎf : y " 5. (a) (b) (c) (d) (e) (f) (g) (h)
f(g(0)) œ f(3) œ 2 g(f(0)) œ g(5) œ 22 f(g(x)) œ f(x# 3) œ x# 3 5 œ x# 2 g(f(x)) œ g(x 5) œ (x 5)# 3 œ x# 10x 22 f(f(5)) œ f(0) œ 5 g(g(2)) œ g(1) œ 2 f(f(x)) œ f(x 5) œ (x 5) 5 œ x 10 g(g(x)) œ g(x# 3) œ (x# 3)# 3 œ x% 6x# 6
6. (a) f ˆg ˆ "# ‰‰ œ f ˆ 23 ‰ œ 3" (b) g ˆf ˆ "# ‰‰ œ g ˆ "# ‰ œ 2 (c) f(g(x)) œ f ˆ x " 1 ‰ œ
" x 1
1œ
(d) g(f(x)) œ g(x 1) œ
" (x1) 1
(e) f(f(2)) œ f(1) œ 0 (f) g(g(2)) œ g ˆ "3 ‰ œ
œ
" 4 3
œ
x x1 " x
3 4
(g) f(f(x)) œ f(x 1) œ (x 1) 1 œ x 2 " (h) g(g(x)) œ g ˆ x " 1 ‰ œ " " 1 œ xx # (x Á 1 and x Á 2) x1
# 7. (a) u(v(f(x))) œ u ˆv ˆ "x ‰‰ œ u ˆ x"# ‰ œ 4 ˆ x" ‰ 5 œ x4# 5 (b) u(f(v(x))) œ u af ax# bb œ u ˆ x"# ‰ œ 4 ˆ x"# ‰ 5 œ x4# 5 # (c) v(u(f(x))) œ v ˆu ˆ "x ‰‰ œ v ˆ4 ˆ x" ‰ 5‰ œ ˆ 4x 5‰
(d) v(f(u(x))) œ v(f(4x 5)) œ v ˆ 4x " 5 ‰ œ ˆ 4x " 5 ‰ (e) f(u(v(x))) œ f au ax# bb œ f a4 ax# b 5b œ
" 4x# 5
(f) f(v(u(x))) œ f(v(4x 5)) œ f a(4x 5)# b œ 8. (a) h(g(f(x))) œ h ˆg ˆÈx‰‰ œ h Š
Èx 4 ‹
#
œ 4Š
" (4x 5)#
Èx 4 ‹
8 œ Èx 8
(b) h(f(g(x))) œ h ˆf ˆ x4 ‰‰ œ h ˆÈ x4 ‰ œ 4È x4 8 œ 2Èx 8 4È x 8 œ Èx 2 4 È4x 8 È œ 4 œ x# 2
(c) g(h(f(x))) œ g ˆh ˆÈx‰‰ œ g ˆ4Èx 8‰ œ (d) g(f(h(x))) œ g(f(4x 8)) œ g ŠÈ4x 8‹
(e) f(g(h(x))) œ f(g(4x 8)) œ f ˆ 4x 4 8 ‰ œ f(x 2) œ Èx 2 (f) f(h(g(x))) œ f ˆh ˆ x ‰‰ œ f ˆ4 ˆ x ‰ 8‰ œ f(x 8) œ Èx 8 4
4
9. (a) y œ f(g(x)) (c) y œ g(g(x)) (e) y œ g(h(f(x)))
(b) y œ j(g(x)) (d) y œ j(j(x)) (f) y œ h(j(f(x)))
10. (a) y œ f(j(x)) (c) y œ h(h(x)) (e) y œ j(g(f(x)))
(b) y œ h(g(x)) œ g(h(x)) (d) y œ f(f(x)) (f) y œ g(f(h(x))) Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
27
28
Chapter 1 Preliminaries g(x)
f(x)
(f ‰ g)(x)
(a)
x7
Èx
Èx 7
(b)
x2
3x
3(x 2) œ 3x 6
(c)
x#
Èx 5
Èx# 5
(d)
x x1
x x1
" x1 " x
1
11.
(e) (f)
x xc1 x xc1 1
" x
(b) af‰gbaxb œ
gaxb" g ax b
œ
x x (x1)
œx
x
" x
12. (a) af‰gbaxb œ lgaxbl œ
œ
x
" lx "l . x x"
Ê"
" g ax b
œ
x x"
Ê"
x x"
œ
" g ax b
Ê
" x"
œ
" gaxb ß so
gaxb œ x ".
(c) Since af‰gbaxb œ Ègaxb œ lxl, gaxb œ x . (d) Since af‰gbaxb œ fˆÈx‰ œ l x l, faxb œ x# . (Note that the domain of the composite is Ò!ß _Ñ.) #
The completed table is shown. Note that the absolute value sign in part (d) is optional. gaxb faxb af‰gbaxb " " lxl x" lx "l x" x
x"
Èx
x# Èx
x#
x x"
lxl lxl
13. (a) fagaxbb œ É 1x 1 œ É 1x x gafaxbb œ
1 Èx 1
(b) Domain af‰gb: Ð0, _Ñ, domain ag‰f b: Ð1, _Ñ (c) Range af‰gb: Ð1, _Ñ, range ag‰f b: Ð0, _Ñ 14. (a) fagaxbb œ 1 2Èx x gafaxbb œ 1 kxk (b) Domain af‰gb: Ð0, _Ñ, domain ag‰f b: Ð0, _Ñ (c) Range af‰gb: Ð0, _Ñ, range ag‰f b: Ð_, 1Ñ 15. (a) y œ (x 7)#
(b) y œ (x 4)#
16. (a) y œ x# 3
(b) y œ x# 5
17. (a) Position 4
(b) Position 1
(c) Position 2
(d) Position 3
18. (a) y œ (x 1)# 4
(b) y œ (x 2)# 3
(c) y œ (x 4)# 1
(d) y œ (x 2)#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 1.5 Combining Functions; Shifting and Scaling Graphs 19.
20.
21.
22.
23.
24.
25.
26.
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29
30
Chapter 1 Preliminaries
27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 1.5 Combining Functions; Shifting and Scaling Graphs 37.
38.
39.
40.
41.
42.
43.
44.
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31
32
Chapter 1 Preliminaries
45.
46.
47.
48.
49. (a) domain: [0ß 2]; range: [#ß $]
(b) domain: [0ß 2]; range: [1ß 0]
(c) domain: [0ß 2]; range: [0ß 2]
(d) domain: [0ß 2]; range: [1ß 0]
(e) domain: [2ß 0]; range: [!ß 1]
(f) domain: [1ß 3]; range: [!ß "]
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 1.5 Combining Functions; Shifting and Scaling Graphs (g) domain: [2ß 0]; range: [!ß "]
(h) domain: [1ß 1]; range: [!ß "]
50. (a) domain: [0ß 4]; range: [3ß 0]
(b) domain: [4ß 0]; range: [!ß $]
(c) domain: [4ß 0]; range: [!ß $]
(d) domain: [4ß 0]; range: ["ß %]
(e) domain: [#ß 4]; range: [3ß 0]
(f) domain: [2ß 2]; range: [3ß 0]
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
33
34
Chapter 1 Preliminaries (g) domain: ["ß 5]; range: [3ß 0]
(h) domain: [0ß 4]; range: [0ß 3]
51. y œ 3x# 3 52. y œ a2xb# 1 œ %x# 1 53. y œ "# ˆ" 54. y œ 1
"‰ x#
" axÎ$b#
œ
" #
œ1
" #x#
* x#
55. y œ È%x 1 56. y œ 3Èx 1 # 57. y œ É% ˆ x# ‰ œ "# È16 x#
58. y œ "$ È% x# 59. y œ " a3xb$ œ " 27x$ $
60. y œ " ˆ x# ‰ œ "
x$ )
"Î# "Î# 61. Let y œ È#x " œ faxb and let gaxb œ x"Î# , haxb œ ˆx "# ‰ , iaxb œ È#ˆx "# ‰ , and "Î# jaxb œ ’È#ˆx "# ‰ “ œ faBb. The graph of haxb is the graph of gaxb shifted left
" #
unit; the graph of iaxb is the graph
of haxb stretched vertically by a factor of È#; and the graph of jaxb œ faxb is the graph of iaxb reflected across the x-axis.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 1.5 Combining Functions; Shifting and Scaling Graphs 62. Let y œ È"
x #
œ faxbÞ Let gaxb œ axb"Î# , haxb œ ax #b"Î# , and iaxb œ
" È # a x
#b"Î# œ È"
x #
35
œ faxbÞ
The graph of gaxb is the graph of y œ Èx reflected across the x-axis. The graph of haxb is the graph of gaxb shifted right two units. And the graph of iaxb is the graph of haxb compressed vertically by a factor of È#.
63. y œ faxb œ x$ . Shift faxb one unit right followed by a shift two units up to get gaxb œ ax "b3 #.
64. y œ a" Bb$ # œ Òax "b$ a#bÓ œ faxb. Let gaxb œ x$ , haxb œ ax "b$ , iaxb œ ax "b$ a#b, and jaxb œ Òax "b$ a#bÓ. The graph of haxb is the graph of gaxb shifted right one unit; the graph of iaxb is the graph of haxb shifted down two units; and the graph of faxb is the graph of iaxb reflected across the x-axis.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
36
Chapter 1 Preliminaries
65. Compress the graph of faxb œ get haxb œ
" #x
".
66. Let faxb œ
" x#
and gaxb œ
# x#
" x
horizontally by a factor of 2 to get gaxb œ
"œ
" # Š B# ‹
"œ
"
#
ŠxÎÈ#‹
"œ
" # ’Š"ÎÈ#‹B“
" #x .
Then shift gaxb vertically down 1 unit to
"Þ Since È# ¸ "Þ%, we see that the graph of
faxb stretched horizontally by a factor of 1.4 and shifted up 1 unit is the graph of gaxb.
$ $ 67. Reflect the graph of y œ faxb œ È x across the x-axis to get gaxb œ È x.
68. y œ faxb œ a#xb#Î$ œ Òa"ba#bxÓ#Î$ œ a"b#Î$ a#xb#Î$ œ a#xb#Î$ . So the graph of faxb is the graph of gaxb œ x#Î$ compressed horizontally by a factor of 2.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 1.5 Combining Functions; Shifting and Scaling Graphs 69.
70.
71. *x# #&y# œ ##& Ê
x# &#
73. $x# ay #b# œ $ Ê
x# "#
75. $ax "b# #ay #b# œ ' Ê
ax " b # #
ŠÈ#‹
y a#b‘# #
ŠÈ$‹
y# $#
a y #b # #
ŠÈ$‹
œ"
74. ax "b# #y# œ % Ê
y# %#
x a"b‘# ##
œ"
# # 76. 'ˆx $# ‰ *ˆy "# ‰ œ &% #
œ"
x# # È Š (‹
72. "'x# (y# œ ""# Ê
œ"
Ê
’xˆ $# ‰“ $#
ˆy "# ‰# #
ŠÈ'‹
œ"
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
y# # ŠÈ#‹
œ"
37
38 77.
Chapter 1 Preliminaries x# "'
y# *
œ " has its center at a!ß !b. Shiftinig 4 units left and 3 units up gives the center at ah, kb œ a%ß $b. So the
equation is
x a4b‘# 4#
ay 3 b # 3#
œ"Ê
ax %b # 4#
a y $b # 3#
œ ". Center, C, is a%ß $b, and major axis, AB, is the segment
from a)ß $b to a!ß $b.
78. The ellipse
x# %
y# #&
œ " has center ah, kb œ a!ß !b. Shifting the ellipse 3 units right and 2 units down produces an ellipse
with center at ah, kb œ a$ß #b and an equation a$ß $b to a$ß (b is the major axis.
ax 3 b# %
y a#b‘# #&
œ ". Center, C, is a3ß #b, and AB, the segment from
79. (a) (fg)(x) œ f(x)g(x) œ f(x)(g(x)) œ (fg)(x), odd (b) Š gf ‹ (x) œ (c) ˆ gf ‰ (x) œ (d) (e) (f) (g) (h) (i)
f(x) g(x) g(x) f(x)
œ œ
f(x) g(x) g(x) f(x)
œ Š gf ‹ (x), odd œ ˆ gf ‰ (x), odd
f # (x) œ f(x)f(x) œ f(x)f(x) œ f # (x), even g# (x) œ (g(x))# œ (g(x))# œ g# (x), even (f ‰ g)(x) œ f(g(x)) œ f(g(x)) œ f(g(x)) œ (f ‰ g)(x), even (g ‰ f)(x) œ g(f(x)) œ g(f(x)) œ (g ‰ f)(x), even (f ‰ f)(x) œ f(f(x)) œ f(f(x)) œ (f ‰ f)(x), even (g ‰ g)(x) œ g(g(x)) œ g(g(x)) œ g(g(x)) œ (g ‰ g)(x), odd
80. Yes, f(x) œ 0 is both even and odd since f(x) œ 0 œ f(x) and f(x) œ 0 œ f(x).
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 1.6 Trigonometric Functions 81. (a)
(b)
(c)
(d)
82.
1.6 TRIGONOMETRIC FUNCTIONS 1. (a) s œ r) œ (10) ˆ 451 ‰ œ 81 m radians and
51 4
1 ‰ 3. ) œ 80° Ê ) œ 80° ˆ 180° œ
41 9
2. ) œ
s r
œ
101 8
œ
51 4
1 ‰ (b) s œ r) œ (10)(110°) ˆ 180° œ
1101 18
œ
551 9
m
ˆ 180° ‰ œ 225° 1 Ê s œ (6) ˆ 491 ‰ œ 8.4 in. (since the diameter œ 12 in. Ê radius œ 6 in.)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
39
40
Chapter 1 Preliminaries
4. d œ 1 meter Ê r œ 50 cm Ê ) œ 5.
1
)
231
0
1 #
s r
œ
30 50
31 4 " È2 È" 2
sin )
0
cos )
1
È #3 "#
tan )
0
È3
0
und.
"
und.
" È3
und.
0
1
und.
È 2
cot )
1
#
und.
È23
sec ) csc )
0
"
"
0
" und.
7. cos x œ 45 , tan x œ 34 9. sin x œ
È8 3
, tan x œ È8
"
‰ ¸ 34° œ 0.6 rad or 0.6 ˆ 180° 1 6.
È2
3#1
) sin )
"
cos )
!
" #
tan )
und.
È 3
cot )
!
È"3
sec )
und.
#
csc )
"
È23
2 È5
10. sin x œ
12 13
11. sin x œ È"5 , cos x œ È25
12. cos x œ
13.
14.
15.
1'
È #3
8. sin x œ
period œ 1
13
, cos x œ
" È2
&1 ' " # È #3
È"3
"
È"3
È 3
"
È 3
2 È3
È2
È23
#
È2
#
"# È3 #
" È5
, tan x œ 12 5 È3 #
, tan x œ
" È3
period œ 41 16.
period œ 2 17.
period œ 4 18.
period œ 6
period œ 1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
1 % " È2
Section 1.6 Trigonometric Functions 19.
20.
period œ 21
period œ 21
21.
22.
period œ 21
period œ 21
23. period œ 1# , symmetric about the origin
24. period œ 1, symmetric about the origin
25. period œ 4, symmetric about the y-axis
26. period œ 41, symmetric about the origin
27. (a) Cos x and sec x are positive in QI and QIV and negative in QII and QIII. Sec x is undefined when cos x is 0. The range of sec x is (_ß 1] ["ß _); the range of cos x is ["ß 1].
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41
42
Chapter 1 Preliminaries (b) Sin x and csc x are positive in QI and QII and negative in QIII and QIV. Csc x is undefined when sin x is 0. The range of csc x is (_ß 1] [1ß _); the range of sin x is ["ß "].
28. Since cot x œ
" tan x
, cot x is undefined when tan x œ 0
and is zero when tan x is undefined. As tan x approaches zero through positive values, cot x approaches infinity. Also, cot x approaches negative infinity as tan x approaches zero through negative values.
29. D: _ x _; R: y œ 1, 0, 1
30. D: _ x _; R: y œ 1, 0, 1
31. cos ˆx 1# ‰ œ cos x cos ˆ 1# ‰ sin x sin ˆ 1# ‰ œ (cos x)(0) (sin x)(1) œ sin x 32. cos ˆx 1# ‰ œ cos x cos ˆ 1# ‰ sin x sin ˆ 1# ‰ œ (cos x)(0) (sin x)(1) œ sin x 33. sin ˆx 1# ‰ œ sin x cos ˆ 1# ‰ cos x sin ˆ 1# ‰ œ (sin x)(0) (cos x)(1) œ cos x 34. sin ˆx 1# ‰ œ sin x cos ˆ 1# ‰ cos x sin ˆ 1# ‰ œ (sin x)(0) (cos x)(1) œ cos x 35. cos (A B) œ cos (A (B)) œ cos A cos (B) sin A sin (B) œ cos A cos B sin A (sin B) œ cos A cos B sin A sin B 36. sin (A B) œ sin (A (B)) œ sin A cos (B) cos A sin (B) œ sin A cos B cos A (sin B) œ sin A cos B cos A sin B 37. If B œ A, A B œ 0 Ê cos (A B) œ cos 0 œ 1. Also cos (A B) œ cos (A A) œ cos A cos A sin A sin A œ cos# A sin# A. Therefore, cos# A sin# A œ 1. 38. If B œ 21, then cos (A 21) œ cos A cos 21 sin A sin 21 œ (cos A)(1) (sin A)(0) œ cos A and sin (A 21) œ sin A cos 21 cos A sin 21 œ (sin A)(1) (cos A)(0) œ sin A. The result agrees with the fact that the cosine and sine functions have period 21. 39. cos (1 x) œ cos 1 cos B sin 1 sin x œ (1)(cos x) (0)(sin x) œ cos x
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 1.6 Trigonometric Functions 40. sin (21 x) œ sin 21 cos (x) cos (21) sin (x) œ (0)(cos (x)) (1)(sin (x)) œ sin x 41. sin ˆ 3#1 x‰ œ sin ˆ 3#1 ‰ cos (x) cos ˆ 3#1 ‰ sin (x) œ (1)(cos x) (0)(sin (x)) œ cos x 42. cos ˆ 3#1 x‰ œ cos ˆ 3#1 ‰ cos x sin ˆ 3#1 ‰ sin x œ (0)(cos x) (1)(sin x) œ sin x œ sin ˆ 14 13 ‰ œ sin
44. cos
111 1#
45. cos
1 12
œ cos ˆ 13 14 ‰ œ cos
46. sin
51 1#
œ sin ˆ 231 14 ‰ œ sin ˆ 231 ‰ cos ˆ 14 ‰ cos ˆ 231 ‰ sin ˆ 14 ‰ œ Š
21 ‰ 3
cos
œ cos
È
47. cos#
1 8
œ
1cos ˆ 281 ‰ #
œ
1 # 2 #
49. sin#
1 1#
œ
1cos ˆ 211# ‰ #
œ
1 # 3 #
È
1 3
1 4
1 3
cos
cos
21 3
1 4
1 3
È2 È3 # ‹Š # ‹
71 1#
œ cos ˆ 14
1 4
È2 ˆ"‰ # ‹ #
43. sin
sin
sin
1 4
cos ˆ 14 ‰ sin
œŠ
sin 1 3
21 3
œŠ
Š
È2 ˆ "‰ # ‹ #
sin ˆ 14 ‰ œ ˆ "# ‰ Š
œ
2 È 2 4
48. cos#
1 1#
œ
2 È 3 4
50. sin#
1 8
Š
È2 # ‹
œ
È2 È3 # ‹Š # ‹
Š
1cos ˆ 211# ‰ #
1cos ˆ 281 ‰ #
51. tan (A B) œ
sin (AB) cos (AB)
œ
sin A cos Bcos A cos B cos A cos Bsin A sin B
œ
sin A cos B cos A sin B cos A cos B cos A cos B sin A sin B cos A cos B cos A cos B cos A cos B
œ
tan Atan B 1tan A tan B
52. tan (A B) œ
sin (AB) cos (AB)
œ
sin A cos Bcos A cos B cos A cos Bsin A sin B
œ
sin A cos B cos A sin B cos A cos B cos A cos B sin A sin B cos A cos B cos A cos B cos A cos B
œ
tan Atan B 1tan A tan B
È 2 È 6 4
œ
È3 È2 # ‹ Š # ‹
È3 È2 # ‹Š # ‹
œ
È 6 È 2 4
œ
ˆ "# ‰ Š
œ
œ
œ
È
1 # 3 #
È
1 # 2 #
œ œ
1 È 3 2È 2 È2 # ‹
œ
1 È 3 2È 2
2 È 3 4
2 È 2 4
53. According to the figure in the text, we have the following: By the law of cosines, c# œ a# b# 2ab cos ) œ 1# 1# 2 cos (A B) œ 2 2 cos (A B). By distance formula, c# œ (cos A cos B)# (sin A sin B)# œ cos# A 2 cos A cos B cos# B sin# A 2 sin A sin B sin# B œ 2 2(cos A cos B sin A sin B). Thus c# œ 2 2 cos (A B) œ 2 2(cos A cos B sin A sin B) Ê cos (A B) œ cos A cos B sin A sin B. 54. (a) cosaA Bb œ cos A cos B sin A sin B sin ) œ cosˆ 1# )‰ and cos ) œ sinˆ 1# )‰ Let ) œ A B
sinaA Bb œ cos’ 1# aA Bb“ œ cos’ˆ 1# A‰ B“ œ cos ˆ 1# A‰ cos B sin ˆ 1# A‰ sin B œ sin A cos B cos A sin B (b) cosaA Bb œ cos A cos B sin A sin B cosaA aBbb œ cos A cos aBb sin A sin aBb Ê cosaA Bb œ cos A cos aBb sin A sin aBb œ cos A cos B sin A asin Bb œ cos A cos B sin A sin B Because the cosine function is even and the sine functions is odd. 55. c# œ a# b# 2ab cos C œ 2# 3# 2(2)(3) cos (60°) œ 4 9 12 cos (60°) œ 13 12 ˆ "# ‰ œ 7. Thus, c œ È7 ¸ 2.65. 56. c# œ a# b# 2ab cos C œ 2# 3# 2(2)(3) cos (40°) œ 13 12 cos (40°). Thus, c œ È13 12 cos 40° ¸ 1.951.
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44
Chapter 1 Preliminaries
57. From the figures in the text, we see that sin B œ hc . If C is an acute angle, then sin C œ hb . On the other hand, if C is obtuse (as in the figure on the right), then sin C œ sin (1 C) œ hb . Thus, in either case, h œ b sin C œ c sin B Ê ah œ ab sin C œ ac sin B. a # b # c # 2ab
By the law of cosines, cos C œ
and cos B œ
a # c # b # . 2ac
Moreover, since the sum of the
interior angles of a triangle is 1, we have sin A œ sin (1 (B C)) œ sin (B C) œ sin B cos C cos B sin C #
#
#
#
#
#
b c c b ˆ h ‰ h ‰ œ ˆ hc ‰ ’ a 2ab a2a# b# c# c# b# b œ “ ’ a 2ac “ b œ ˆ 2abc
ah bc
Ê ah œ bc sin A.
Combining our results we have ah œ ab sin C, ah œ ac sin B, and ah œ bc sin A. Dividing by abc gives h sin A sin C sin B bc œ ðóóóóóóóñóóóóóóóò a œ c œ b . law of sines 58. By the law of sines, Thus sin B œ
3È 3 2È 7
sin A #
œ
sin B 3
œ
È3/2 c .
By Exercise 55 we know that c œ È7.
¶ 0.982.
59. From the figure at the right and the law of cosines, b# œ a# 2# 2(2a) cos B œ a# 4 4a ˆ "# ‰ œ a# 2a 4. Applying the law of sines to the figure, Ê
È2/2 a
œ
È3/2 b
sin A a
œ
sin B b
Ê b œ É 3# a. Thus, combining results,
a# 2a 4 œ b# œ
3 #
a# Ê 0 œ
" #
a# 2a 4
Ê 0 œ a# 4a 8. From the quadratic formula and the fact that a 0, we have aœ
4È4# 4(1)(8) #
œ
4 È 3 4 #
¶ 1.464.
60. (a) The graphs of y œ sin x and y œ x nearly coincide when x is near the origin (when the calculator is in radians mode). (b) In degree mode, when x is near zero degrees the sine of x is much closer to zero than x itself. The curves look like intersecting straight lines near the origin when the calculator is in degree mode. 61. A œ 2, B œ 21, C œ 1, D œ 1
62. A œ "# , B œ 2, C œ 1, D œ
" #
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Section 1.6 Trigonometric Functions 63. A œ 12 , B œ 4, C œ 0, D œ
64. A œ
L 21 ,
" 1
B œ L, C œ 0, D œ 0
65. (a) amplitude œ kAk œ 37 (c) right horizontal shift œ C œ 101
(b) period œ kBk œ 365 (d) upward vertical shift œ D œ 25
66. (a) It is highest when the value of the sine is 1 at f(101) œ 37 sin (0) 25 œ 62° F. The lowest mean daily temp is 37(1) 25 œ 12° F. (b) The average of the highest and lowest mean daily temperatures œ The average of the sine function is its horizontal axis, y œ 25.
62°(12)° #
œ 25° F.
67-70. Example CAS commands: Maple f := x -> A*sin((2*Pi/B)*(x-C))+D1; A:=3; C:=0; D1:=0; f_list := [seq( f(x), B=[1,3,2*Pi,5*Pi] )]; plot( f_list, x=-4*Pi..4*Pi, scaling=constrained, color=[red,blue,green,cyan], linestyle=[1,3,4,7], legend=["B=1","B=3","B=2*Pi","B=3*Pi"], title="#67 (Section 1.6)" ); Mathematica Clear[a, b, c, d, f, x] f[x_]:=a Sin[21/b (x c)] + d Plot[f[x]/.{a Ä 3, b Ä 1, c Ä 0, d Ä 0}, {x, 41, 41 }] 67. (a) The graph stretches horizontally.
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45
46
Chapter 1 Preliminaries (b) The period remains the same: period œ l B l. The graph has a horizontal shift of
" #
period.
68. (a) The graph is shifted right C units.
(b) The graph is shifted left C units. (c) A shift of „ one period will produce no apparent shift. l C l œ ' 69. The graph shifts upwards l D lunits for D ! and down l D lunits for D !Þ
70. (a) The graph stretches l A l units.
(b) For A !, the graph is inverted. 1.7 GRAPHING WITH CALCULATORS AND COMPUTERS 1-4.
The most appropriate viewing window displays the maxima, minima, intercepts, and end behavior of the graphs and has little unused space.
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Section 1.7 Graphing with Calculators and Computers 1. d.
2. c.
3. d.
4. b.
5-30.
For any display there are many appropriate display widows. The graphs given as answers in Exercises 530 are not unique in appearance.
5. Ò2ß 5Ó by Ò15ß 40Ó
6. Ò4ß 4Ó by Ò4ß 4Ó
7. Ò2ß 6Ó by Ò250ß 50Ó
8. Ò1ß 5Ó by Ò5ß 30Ó
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47
48
Chapter 1 Preliminaries
9. Ò4ß 4Ó by Ò5ß 5Ó
10. Ò2ß 2Ó by Ò2ß 8Ó
11. Ò2ß 6Ó by Ò5ß 4Ó
12. Ò4ß 4Ó by Ò8ß 8Ó
13. Ò"ß 'Ó by Ò"ß %Ó
14. Ò"ß 'Ó by Ò"ß &Ó
15. Ò3ß 3Ó by Ò!ß "!Ó
16. Ò"ß #Ó by Ò!ß "Ó
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Section 1.7 Graphing with Calculators and Computers 17. Ò&ß "Ó by Ò&ß &Ó
18. Ò&ß "Ó by Ò#ß %Ó
19. Ò%ß %Ó by Ò!ß $Ó
20. Ò&ß &Ó by Ò#ß #Ó
21. Ò"!ß "!Ó by Ò'ß 'Ó
22. Ò&ß &Ó by Ò#ß #Ó
23. Ò'ß "!Ó by Ò'ß 'Ó
24. Ò$ß &Ó by Ò#ß "!Ó
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49
50
Chapter 1 Preliminaries
25. Ò!Þ!$ß !Þ!$Ó by Ò"Þ#&ß "Þ#&Ó
26. Ò!Þ"ß !Þ"Ó by Ò$ß $Ó
27. Ò$!!ß $!!Ó by Ò"Þ#&ß "Þ#&Ó
28. Ò&!ß &!Ó by Ò!Þ"ß !Þ"Ó
29. Ò!Þ#&ß !Þ#&Ó by Ò!Þ$ß !Þ$Ó
30. Ò!Þ"&ß !Þ"&Ó by Ò!Þ!#ß !Þ!&Ó
31. x# #x œ % %y y# Ê y œ # „ Èx# #x ). The lower half is produced by graphing y œ # Èx# #x ).
32. y# "'x# œ " Ê y œ „ È" "'x# . The upper branch is produced by graphing y œ È" "'x# .
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Section 1.7 Graphing with Calculators and Computers 33.
34.
35.
36.
37.
38Þ
39.
40.
41. (a) y œ "!&*Þ"%x #!(%*(# (b) m œ "!&*Þ"% dollars/year, which is the yearly increase in compensation.
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51
52
Chapter 1 Preliminaries (c)
(d) Answers may vary slightly. y œ a"!&*Þ14ba#!"!b #!(%*(# œ $&$ß 899 42. (a) Let C œ cost and x œ year. C œ a(*'!Þ("bx "Þ' ‚ "!( (b) Slope represents increase in cost per year (c) C œ a#'$(Þ"%bx &Þ# ‚ "!' (d) The median price is rising faster in the northeast (the slope is larger). 43. (a) Let x represent the speed in miles per hour and d the stopping distance in feet. The quadratic regression function is d œ !Þ!)''x# "Þ*(x &!Þ". (b)
(c) From the graph in part (b), the stopping distance is about $(! feet when the vehicle is (# mph and it is about &#& feet when the speed is )& mph. Algebraically: dquadratic a(#b œ !Þ!)''a(#b# "Þ*(a(#b &!Þ" œ $'(Þ' ft. dquadratic a)&b œ !Þ!)''a)&b# "Þ*(a)&b &!Þ" œ &##Þ) ft. (d) The linear regression function is d œ 'Þ)*x "%!Þ% Ê dlinear a(#b œ 'Þ)*a(#b "%!Þ% œ $&&Þ( ft and dlinear a)&b œ 'Þ)*a)&b "%!Þ% œ %%&Þ# ft. The linear regression line is shown on the graph in part (b). The quadratic regression curve clearly gives the better fit.
44. (a) The power regression function is y œ %Þ%%'%(x!Þ&""%"% .
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 1 Practice Exercises (b)
(c) 15Þ2 km/h (d) The linear regression function is y œ !Þ*"$'(&x %Þ")**(' and it is shown on the graph in part (b). The linear regession function gives a speed of "%Þ# km/h when y œ "" m. The power regression curve in part (a) better fits the data. CHAPTER 1 PRACTICE EXERCISES 1. ( 2x $ Ê #x % Ê x # 2.
3x "! Ê x "! $
3.
" & ax
"b "% ax #b Ê %ax "b &ax #b
x$ #
%$ x Ê $ax $b #a% xb
qqqqqqqqðïïïïïïïî x "! $
Ê %x % &x "! Ê ' x
4.
Ê $x * ) #x Ê &x " Ê x 5.
qqqqqqqqñïïïïïïïî x " &
" &
lx " l œ ( Ê x " œ ( or ax "b œ ( Ê x œ ' or x œ )
6. ly $ l % Ê % y $ % Ê " y ( 7. ¹" x# ¹
$ #
Ê "
x #
$# or "
x #
$ #
Ê x# &# or x#
" #
Ê x & or x "
Ê x & or x " 8. ¹ #x$( ¹ Ÿ & Ê & Ÿ
#x( $
Ÿ & Ê 1& Ÿ #x ( Ÿ 1& Ê 22 Ÿ #x Ÿ 8 Ê "" Ÿ x Ÿ %
9. Since the particle moved to the y-axis, # ?x œ ! Ê ?x œ 2. Since ?y œ 3?x œ 6, the new coordinates are (x ?xß y ?y) œ (# #ß & ') œ (0ß 11). 10. (a)
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53
54
Chapter 1 Preliminaries (b)
line AB
slope 10 1 9 3 2 8 œ 6 œ # 10 6 4 2 2 (4) œ 6 œ 3 6 (3) 9 3 % 2 œ 6 œ # 1 (3) 4 2 82 œ 6 œ 3 66 œ0 % 14 3
BC CD DA CE
BD is vertical and has no slope (c) Yes; A, B, C and D form a parallelogram. 3 ˆ 14 ‰ (d) Yes. The line AB has equation y 1 œ 3# (x 8). Replacing x by 14 3 gives y œ # 3 8 " 3 ˆ 10 ‰ 14 œ # 3 1 œ 5 1 œ 6. Thus, E ˆ 3 ß 6‰ lies on the line AB and the points A, B and E are collinear. (e) The line CD has equation y 3 œ 3# (x 2) or y œ 3# x. Thus the line passes through the origin. 11. The triangle ABC is neither an isosceles triangle nor is it a right triangle. The lengths of AB, BC and AC are È53, È72 and È65, respectively. The slopes of AB, BC and AC are 7 , 1 and " , respectively. #
8
12. P(xß 3x 1) is a point on the line y œ 3x 1. If the distance from P to (!ß 0) equals the distance from P to ($ß %), then x# (3x 1)# œ (x 3)# (3 3x)# Ê x# 9x# 6x 1 œ x# 6x 9 9 18x 9x# 23 ˆ 17 ‰ ˆ 17 23 ‰ Ê 18x œ 17 or x œ 17 18 Ê y œ 3x 1 œ 3 18 1 œ 6 . Thus the point is P 18 ß 6 . 13. y œ $ax "b a'b Ê y œ $x * 14. y œ "# ax "b # Ê y œ "# x
$ #
15. x œ ! 16. m œ
# ' " a$b
œ
) %
œ # Ê y œ #ax $b ' Ê y œ #x
17. y œ # 18. m œ
&$ # $
œ
# &
œ &# Ê y œ &# ax $b $ Ê y œ &# x
#" &
19. y œ $x $ 20. Since #x y œ # is equivalent to y œ #x #, the slope of the given line (and hence the slope of the desired line) is 2. y œ #a x "b " Ê y œ # x & 21. Since %x $y œ "# is equivalent to y œ %$ x %, the slope of the given line (and hence the slope of the desired line) is %$ . y œ %$ ax 4b "2 Ê y œ %$ x
#! $
22. Since $x &y œ " is equivalent to y œ $& x "& , the slope of the given line is 5$ .
yœ
5$ ax
#b $ Ê y œ
5$ x
"* $
$ &
and the slope of the perpendicular line is
23. Since "# x "$ y œ " is equivalent to y œ $# x $, the slope of the given line is $# and the slope of the perpendicular line is #$ . y œ #$ ax "b # Ê y œ #$ x
) $
24. The line passes through a!ß &b and a$ß !b. m œ
! a&b $!
œ
& $
Ê y œ $& x &
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Chapter 1 Practice Exercises 25. The area is A œ 1 r# and the circumference is C œ #1 r. Thus, r œ 26. The surface area is S œ %1 r# Ê r œ ˆ %S1 ‰ surface area gives S œ %1 r# œ %1 ˆ $%V1 ‰
"Î#
#Î$
C #1
#
Ê A œ 1ˆ #C1 ‰ œ
C# %1 .
$ $V . The volume is V œ %$ 1 r$ Ê r œ É %1 . Substitution into the formula for
.
27. The coordinates of a point on the parabola are axß x# b. The angle of inclination ) joining this point to the origin satisfies the equation tan ) œ 28. tan ) œ
rise run
œ
h &!!
x# x
œ x. Thus the point has coordinates axß x# b œ atan )ß tan# )b.
Ê h œ &!! tan ) ft.
29.
30.
Symmetric about the origin.
Symmetric about the y-axis.
31.
32.
Neither
Symmetric about the y-axis.
33. yaxb œ axb# " œ x# " œ yaxb. Even. 34. yaxb œ axb& axb$ axb œ x& x$ x œ yaxb. Odd. 35. yaxb œ " cosaxb œ " cos x œ yaxb. Even. 36. yaxb œ secaxb tanaxb œ 37. yaxb œ
axb% " axb$ #axb
œ
x% " x$ #x
sinaxb cos# axb
œ
sin x cos# x
œ sec x tan x œ yaxb. Odd.
%
" œ xx$ # x œ yaxb. Odd.
38. yaxb œ " sinaxb œ " sin x. Neither even nor odd. 39. yaxb œ x cosaxb œ x cos x. Neither even nor odd. 40. yaxb œ Éaxb% " œ Èx% " œ yaxb. Even.
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Chapter 1 Preliminaries
41. (a) The function is defined for all values of x, so the domain is a_ß _b. (b) Since l x l attains all nonnegative values, the range is Ò#ß _Ñ. 42. (a) Since the square root requires " x !, the domain is Ð_ß "Ó. (b) Since È" x attains all nonnegative values, the range is Ò#ß _Ñ. 43. (a) Since the square root requires "' x# !, the domain is Ò%ß %Ó. (b) For values of x in the domain, ! Ÿ "' x# Ÿ "', so ! Ÿ È"' x# Ÿ %. The range is Ò!ß %Ó. 44. (a) The function is defined for all values of x, so the domain is a_ß _b. (b) Since $#x attains all positive values, the range is a"ß _b. 45. (a) The function is defined for all values of x, so the domain is a_ß _b. (b) Since #ex attains all positive values, the range is a$ß _b. 46. (a) The function is equivalent to y œ tan #x, so we require #x Á
k1 #
for odd integers k. The domain is given by x Á
k1 %
for
odd integers k. (b) Since the tangent function attains all values, the range is a_ß _b. 47. (a) The function is defined for all values of x, so the domain is a_ß _b. (b) The sine function attains values from " to ", so # Ÿ #sina$x 1b Ÿ # and hence $ Ÿ #sina$x 1b " Ÿ ". The range is Ò3ß 1Ó. 48. (a) The function is defined for all values of x, so the domain is a_ß _b. & (b) The function is equivalent to y œ È x# , which attains all nonnegative values. The range is Ò!ß _Ñ. 49. (a) The logarithm requires x $ !, so the domain is a$ß _b. (b) The logarithm attains all real values, so the range is a_ß _b. 50. (a) The function is defined for all values of x, so the domain is a_ß _b. (b) The cube root attains all real values, so the range is a_ß _b. 51. (a) The function is defined for % Ÿ x Ÿ %, so the domain is Ò%ß %Ó. (b) The function is equivalent to y œ Èl x l, % Ÿ x Ÿ %, which attains values from ! to # for x in the domain. The range is Ò!ß #Ó. 52. (a) The function is defined for # Ÿ x Ÿ #, so the domain is Ò#ß #Ó. (b) The range is Ò"ß "Ó. 53. First piece: Line through a!ß "b and a"ß !b. m œ Second piece: Line through a"ß "b and a#ß !b. m faxb œ œ
" x, ! Ÿ x " # x, " Ÿ x Ÿ #
54. First piece: Line through a!ß !b and a2ß 5b. m œ Second piece: Line through a2ß 5b and a4ß !b. m faxb œ
10
5 2 x, 5x 2 ,
!" " "! œ " œ " " œ !# " œ "
" Ê y œ x " œ " x œ " Ê y œ ax "b " œ x # œ # x
5! 5 5 2! œ 2 Ê y œ 2x 5 5 5 œ !4 2 œ 2 œ 2 Ê
y œ 52 ax 2b 5 œ 52 x 10 œ 10
!Ÿx2 (Note: x œ 2 can be included on either piece.) 2ŸxŸ4 Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
5x 2
Chapter 1 Practice Exercises 55. (a) af‰gba"b œ faga"bb œ fŠ È"" # ‹ œ fa"b œ (b) ag‰f ba#b œ gafa#bb œ gˆ "2 ‰ œ (c) af‰f baxb œ fafaxbb œ fˆ "x ‰ œ
" É "# #
" "Îx
œ
" È#Þ&
" "
œ"
or É &#
œ x, x Á !
(d) ag‰gbaxb œ gagaxbb œ gŠ Èx" # ‹ œ
" " É Èx # #
œ
% x# È É " #È x #
$ 56. (a) af‰gba"b œ faga"bb œ fˆÈ " "‰ œ fa!b œ # ! œ # $ (b) ag‰f ba#b œ faga#bb œ ga# #b œ ga!b œ È !"œ"
(c) af‰f baxb œ fafaxbb œ fa# xb œ # a# xb œ x $ $ $ È (d) ag‰gbaxb œ gagaxbb œ gˆÈ x "‰ œ É x""
#
57. (a) af‰gbaxb œ fagaxbb œ fˆÈx #‰ œ # ˆÈx #‰ œ x, x #. ag‰f baxb œ fagaxbb œ ga# x# b œ Èa# x# b # œ È% x# (b) Domain of f‰g: Ò#ß _ÑÞ Domain of g‰f: Ò#ß #ÓÞ (c) Range of f‰g: Ð_ß #ÓÞ Range of g‰f: Ò!ß #ÓÞ % 58. (a) af‰gbaxb œ fagaxbb œ fŠÈ" x‹ œ ÉÈ" x œ È " x.
ag‰f baxb œ fagaxbb œ gˆÈx‰ œ É" Èx (b) Domain of f‰g: Ð_ß "ÓÞ Domain of g‰f: Ò!ß "ÓÞ (c) Range of f‰g: Ò!ß _ÑÞ Range of g‰f: Ò!ß "ÓÞ 59.
60.
The graph of f# (x) œ f" akxkb is the same as the graph of f" (x) to the right of the y-axis. The graph of f# (x) to the left of the y-axis is the reflection of y œ f" (x), x 0 across the y-axis.
The graph of f# (x) œ f" akxkb is the same as the graph of f" (x) to the right of the y-axis. The graph of f# (x) to the left of the y-axis is the reflection of y œ f" (x), x 0 across the y-axis.
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57
58
Chapter 1 Preliminaries
61.
62.
The graph of f# (x) œ f" akxkb is the same as the graph of f" (x) to the right of the y-axis. The graph of f# (x) to the left of the y-axis is the reflection of y œ f" (x), x 0 across the y-axis.
It does not change the graph.
63.
64.
The graph of f# (x) œ f" akxkb is the same as the graph of f" (x) to the right of the y-axis. The graph of f# (x) to the left of the y-axis is the reflection of y œ f" (x), x 0 across the y-axis. 65.
66.
Whenever g" (x) is positive, the graph of y œ g# (x) œ kg" (x)k is the same as the graph of y œ g" (x). When g" (x) is negative, the graph of y œ g# (x) is the reflection of the graph of y œ g" (x) across the x-axis. 67.
The graph of f# (x) œ f" akxkb is the same as the graph of f" (x) to the right of the y-axis. The graph of f# (x) to the left of the y-axis is the reflection of y œ f" (x), x 0 across the y-axis.
It does not change the graph.
Whenever g" (x) is positive, the graph of y œ g# (x) œ kg" (x)k is the same as the graph of y œ g" (x). When g" (x) is negative, the graph of y œ g# (x) is the reflection of the graph of y œ g" (x) across the x-axis.
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Chapter 1 Practice Exercises
59
68.
Whenever g" (x) is positive, the graph of y œ g# (x) œ kg" (x)k is the same as the graph of y œ g" (x). When g" (x) is negative, the graph of y œ g# (x) is the reflection of the graph of y œ g" (x) across the x-axis.
69.
70.
period œ 1
period œ 41
71.
72.
period œ 2
period œ 4
73.
74.
period œ 21
period œ 21
75. (a) sin B œ sin
1 3
œ
b c
œ
b #
Ê b œ 2 sin
1 3
œ 2Š
È3 # ‹
œ È3. By the theorem of Pythagoras,
a# b# œ c# Ê a œ Èc# b# œ È4 3 œ 1. (b) sin B œ sin
1 3
œ
b c
œ
2 c
Ê cœ
2 sin 13
œ È23 œ Š ‹ #
4 È3
#
. Thus, a œ Èc# b# œ ÊŠ È43 ‹ (2)# œ É 43 œ
76. (a) sin A œ
a c
Ê a œ c sin A
(b) tan A œ
a b
Ê a œ b tan A
77. (a) tan B œ
b a
Ê aœ
(b) sin A œ
a c
Ê cœ
b tan B
a sin A
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
2 È3
.
60
Chapter 1 Preliminaries
78. (a) sin A œ
(c) sin A œ
a c
a c
œ
È c # b # c
79. Let h œ height of vertical pole, and let b and c denote the distances of points B and C from the base of the pole, measured along the flatground, respectively. Then, tan 50° œ hc , tan 35° œ hb , and b c œ 10. Thus, h œ c tan 50° and h œ b tan 35° œ (c 10) tan 35° Ê c tan 50° œ (c 10) tan 35° Ê c (tan 50° tan 35°) œ 10 tan 35° tan 35° Ê c œ tan10 50°tan 35° Ê h œ c tan 50° œ
10 tan 35° tan 50° tan 50°tan 35°
¸ 16.98 m.
80. Let h œ height of balloon above ground. From the figure at the right, tan 40° œ ha , tan 70° œ hb , and a b œ 2. Thus, h œ b tan 70° Ê h œ (2 a) tan 70° and h œ a tan 40° Ê (2 a) tan 70° œ a tan 40° Ê a(tan 40° tan 70°) 70° œ 2 tan 70° Ê a œ tan 240°tantan 70° Ê h œ a tan 40° œ
2 tan 70° tan 40° tan 40°tan 70°
¸ 1.3 km.
81. (a)
(b) The period appears to be 41. (c) f(x 41) œ sin (x 41) cos ˆ x#41 ‰ œ sin (x 21) cos ˆ x# 21‰ œ sin x cos since the period of sine and cosine is 21. Thus, f(x) has period 41.
x #
82. (a)
(b) D œ (_ß 0) (!ß _); R œ [1ß 1] (c) f is not periodic. For suppose f has period p. Then f ˆ #"1 kp‰ œ f ˆ #"1 ‰ œ sin 21 œ 0 for all integers k. Choose k so large that
" #1
kp
" 1
Ê 0
" (1/21)kp
1. But then
f ˆ #"1 kp‰ œ sin Š (1/#1")kp ‹ 0 which is a contradiction. Thus f has no period, as claimed.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 1 Additional and Advanced Exercises CHAPTER 1 ADDITIONAL AND ADVANCED EXERCISES 1. (a) The given graph is reflected about the y-axis.
(b) The given graph is reflected about the x-axis.
(c) The given graph is shifted left 1 unit, stretched vertically by a factor of 2, reflected about the x-axis, and then shifted upward 1 unit.
2. (a)
(d) The given graph is shifted right 2 units, stretched vertically by a factor of 3, and then shifted downward 2 units.
(b)
3. There are (infinitely) many such function pairs. For example, f(x) œ 3x and g(x) œ 4x satisfy f(g(x)) œ f(4x) œ 3(4x) œ 12x œ 4(3x) œ g(3x) œ g(f(x)). 4. Yes, there are many such function pairs. For example, if g(x) œ (2x 3)$ and f(x) œ x"Î$ , then (f ‰ g)(x) œ f(g(x)) œ f a(2x 3)$ b œ a(2x 3)$ b
"Î$
œ 2x 3.
5. If f is odd and defined at x, then f(x) œ f(x). Thus g(x) œ f(x) 2 œ f(x) 2 whereas g(x) œ (f(x) 2) œ f(x) 2. Then g cannot be odd because g(x) œ g(x) Ê f(x) 2 œ f(x) 2 Ê 4 œ 0, which is a contradiction. Also, g(x) is not even unless f(x) œ 0 for all x. On the other hand, if f is even, then g(x) œ f(x) 2 is also even: g(x) œ f(x) 2 œ f(x) 2 œ g(x). 6. If g is odd and g(0) is defined, then g(0) œ g(0) œ g(0). Therefore, 2g(0) œ 0 Ê g(0) œ 0.
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62
Chapter 1 Preliminaries
7. For (xß y) in the 1st quadrant, kxk kyk œ 1 x Í x y œ 1 x Í y œ 1. For (xß y) in the 2nd quadrant, kxk kyk œ x 1 Í x y œ x 1 Í y œ 2x 1. In the 3rd quadrant, kxk kyk œ x 1 Í x y œ x 1 Í y œ 2x 1. In the 4th quadrant, kxk kyk œ x 1 Í x (y) œ x 1 Í y œ 1. The graph is given at the right. 8. We use reasoning similar to Exercise 7. (1) 1st quadrant: y kyk œ x kxk Í 2y œ 2x Í y œ x. (2) 2nd quadrant: y kyk œ x kxk Í 2y œ x (x) œ 0 Í y œ 0. (3) 3rd quadrant: y kyk œ x kxk Í y (y) œ x (x) Í 0 œ 0 Ê all points in the 3rd quadrant satisfy the equation. (4) 4th quadrant: y kyk œ x kxk Í y (y) œ 2x Í 0 œ x. Combining these results we have the graph given at the right:
9. By the law of sines, 10. By the law of sines,
sin 13 È3
œ
sin 14 4
œ
sin A a
œ
sin A a
œ
sin B b
œ
sin B b
œ
sin 14 b
Ê bœ
sin B 3
Ê sin B œ
È3 sin (1/4) sin (1/3) 3 4
11. By the law of cosines, a# œ b# c# 2bc cos A Ê cos A œ
sin
œ
È3 Š È2 ‹
È3
#
œ È2.
#
1 4
œ
b # c # a# 2bc
3 4
Š
œ
12. By the law of cosines, c# œ a# b# 2ab cos C œ 2# 3# (2)(2)(3) cos
È2 # ‹
œ
3È 2 8
2# 3# 2# 2(2)(3)
1 4
.
œ 34 .
œ 4 9 12 Š
È2 # ‹
œ 13 6È2 Ê c œ É13 6È2 , since c 0. # a # c # b # 4 # 3 # œ 2(2)(2)(4) #ac È135 3È15 121 256 œ 16 œ 16 .
œ
4169 16
# 4 # 5 # a # b # c # œ 2(2)(2)(4) 2ab È231 25 256 œ 16 .
œ
41625 16
13. By the law of cosines, b# œ a# c# 2ac cos B Ê cos B œ œ
11 16 .
Since 0 B 1, sin B œ È1 cos# B œ É1
14. By the law of cosines, c# œ a# b# 2ab cos C Ê cos C œ 5 œ 16 . Since 0 C 1, sin C œ È1 cos# C œ É1
15. (a) sin# x cos# x œ 1 Ê sin# x œ 1 cos# x œ (1 cos x)(1 cos x) Ê (1 cos x) œ Ê
1cos x sin x
œ
sin# x 1cos x
sin x 1cos x
(b) Using the definition of the tangent function and the double angle formulas, we have tan# ˆ x# ‰ œ
sin# ˆ x# ‰ cos# ˆ #x ‰
œ
"cos Š2 Š #x ‹‹ #
"cos Š2 Š #x ‹‹ #
œ
1cos x 1cos x
.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 1 Additional and Advanced Exercises 16. The angles labeled # in the accompanying figure are equal since both angles subtend arc CD. Similarly, the two angles labeled ! are equal since they both subtend arc AB. Thus, triangles AED and BEC are similar which ) b implies ab c œ 2a cos a c Ê (a c)(a c) œ b(2a cos ) b) Ê a# c# œ 2ab cos ) b# Ê c# œ a# b# 2ab cos ).
17. As in the proof of the law of sines of Section P.5, Exercise 57, ah œ bc sin A œ ab sin C œ ac sin B Ê the area of ABC œ "# (base)(height) œ "# ah œ "# bc sin A œ "# ab sin C œ "# ac sin B. 18. As in Section P.5, Exercise 57, (Area of ABC)# œ œ
" 4
(base)# (height)# œ
" 4
a# h # œ
" 4
a# b# sin# C
a# b# a" cos# Cb . By the law of cosines, c# œ a# b# 2ab cos C Ê cos C œ
Thus, (area of ABC)# œ œ
" 4
" 16
" 4
a# b# a" cos# Cb œ #
Š4a# b# aa# b# c# b ‹ œ
" 16
" 4
a# b# Œ" Š a
#
b # c# ‹ #ab
#
œ
a# b# 4
a # b # c# 2ab
Š"
#
.
#
aa b c # b 4a# b#
#
‹
ca2ab aa# b# c# bb a2ab aa# b# c# bbd
" ca(a b)# c# b ac# (a b)# bd œ 16 c((a b) c)((a b) c)(c (a b))(c (a b))d a b c a b c a b c a b c œ ˆ # ‰ ˆ # ‰ ˆ # ‰ ˆ # ‰‘ œ s(s a)(s b)(s c), where s œ a#bc .
œ
" 16
Therefore, the area of ABC equals Ès(s a)(s b)(s c) . 19. 1. 2. 3. 4. 5. 6.
b c (a c) œ b a, which is positive since a b. Thus, a c b c. b c (a c) œ b a, which is positive since a b. Thus, a c b c. c 0 and a b Ê c 0 œ c and b a are positive Ê (b a)c œ bc ac is positive Ê ac bc. a b and c 0 Ê b a and c are positive Ê (b a)(c) œ ac bc is positive Ê bc ac. Since a 0, a and "a are positive Ê "a 0. Since 0 a b, both "a and b" are positive. By (3), a b and "a 0 Ê a ˆ "a ‰ b ˆ "a ‰ or 1 ba Ê 1 ˆ "b ‰
7.
b a
b a
" a
" b
0 Ê
" b
" a . " a
and b" are both negative, i.e., 0 and b" 0. By (4), a b and 1 Ê 1 ˆ b" ‰ ba ˆ b" ‰ by (4) since b" 0 Ê b" "a .
ab0 Ê Ê
ˆ b" ‰ by (3) since
" a
0 Ê b ˆ "a ‰ a ˆ "a ‰
20. (a) If a œ 0, then 0 œ kak kbk Í b Á 0 Í 0 œ kak# kbk# . Since kak# œ kak kak œ ka# k œ a# and kbk# œ b# we obtain a# b# . If a Á 0 then kak 0 and kak kbk Ê a# b# . On the other hand, if a# b# then a# œ kak# kbk# œ b# Ê 0 kbk# kak# œ akbk kakb akbk kakb . Since akbk kakb 0 and the product akbk kakb akbk kakb is positive, we must have akbk kakb 0 Ê kbk kak . Thus kak kbk Í a# b# . (b) ab Ÿ kabk Ê ab 2 kabk by Exercise 19(4) above Ê a# 2ab b# kak# 2 kak kbk kbk# , since kak# œ a# and kbk# œ b# . Factoring both sides, (a b)# akak kbkb# Ê ka bk kkak kbkk , by part (a). 21. The fact that ka" a# á an k Ÿ ka" k ka# k á kan k holds for n œ 1 is obvious. It also holds for n œ 2 by the triangle inequality. We now show it holds for all positive integers n, by induction. Suppose it holds for n œ k 1: ka" a# á ak k Ÿ ka" k ka# k á kak k (this is the induction hypothesis). Then ka" a# á ak akb1 k œ kaa" a# á ak b akb1 k Ÿ ka" a# á ak k kakb1 k (by the triangle inequality) Ÿ ka" k ka# k á kak k kakb1 k (by the induction hypothesis) and the inequality holds for n œ k 1. Hence it holds for all n by induction. Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
63
64
Chapter 1 Preliminaries
22. The fact that ka" a# á an k ka" k ka# k á kan k holds for n œ 1 is obvious. It holds for n œ 2 by Exercise 21(b), since ka" a# k œ ka" (a# )k kka" k ka# kk œ kka" k ka# kk ka" k ka# k . We now show it holds for all positive integers n by induction. Suppose the inequality holds for n œ k 1. Then ka" a# á ak k ka" k ka# k á kak k (this is the induction hypothesis). Thus ka" á ak akb1 k œ kaa" á ak b aakb1 bk kkaa" á ak bk kakb1 kk (by Exercise 21(b)) œ kka" á ak k kakb1 kk ka" á ak k kakb1 k ka" k ka# k á kak k kakb1 k (by the induction hypothesis). Hence the inequality holds for all n by induction. 23. If f is even and odd, then f(x) œ f(x) and f(x) œ f(x) Ê f(x) œ f(x) for all x in the domain of f. Thus 2f(x) œ 0 Ê f(x) œ 0. f(x) f((x)) œ f(x) #f(x) œ E(x) Ê E # even function. Define O(x) œ f(x) E(x) œ f(x) f(x) #f(x) œ f(x) #f(x) . Then O(x) œ f(x) #f((x)) œ f(x)# f(x) œ Š f(x) #f(x) ‹ œ O(x) Ê O is an odd function
24. (a) As suggested, let E(x) œ
f(x) f(x) #
Ê E(x) œ
is an
Ê f(x) œ E(x) O(x) is the sum of an even and an odd function. (b) Part (a) shows that f(x) œ E(x) O(x) is the sum of an even and an odd function. If also f(x) œ E" (x) O" (x), where E" is even and O" is odd, then f(x) f(x) œ 0 œ aE" (x) O" (x)b (E(x) O(x)). Thus, E(x) E" (x) œ O" (x) O(x) for all x in the domain of f (which is the same as the domain of E E" and O O" ). Now (E E" )(x) œ E(x) E" (x) œ E(x) E" (x) (since E and E" are even) œ (E E" )(x) Ê E E" is even. Likewise, (O" O)(x) œ O" (x) O(x) œ O" (x) (O(x)) (since O and O" are odd) œ (O" (x) O(x)) œ (O" O)(x) Ê O" O is odd. Therefore, E E" and O" O are both even and odd so they must be zero at each x in the domain of f by Exercise 23. That is, E" œ E and O" œ O, so the decomposition of f found in part (a) is unique. 25. y œ ax# bx c œ a Šx# ba x
b# 4a# ‹
b# 4a
c œ a ˆx
b ‰# 2a
b# 4a
c
(a) If a 0 the graph is a parabola that opens upward. Increasing a causes a vertical stretching and a shift of the vertex toward the y-axis and upward. If a 0 the graph is a parabola that opens downward. Decreasing a causes a vertical stretching and a shift of the vertex toward the y-axis and downward. (b) If a 0 the graph is a parabola that opens upward. If also b 0, then increasing b causes a shift of the graph downward to the left; if b 0, then decreasing b causes a shift of the graph downward and to the right. If a 0 the graph is a parabola that opens downward. If b 0, increasing b shifts the graph upward to the right. If b 0, decreasing b shifts the graph upward to the left. (c) Changing c (for fixed a and b) by ?c shifts the graph upward ?c units if ?c 0, and downward ?c units if ?c 0. 26. (a) If a 0, the graph rises to the right of the vertical line x œ b and falls to the left. If a 0, the graph falls to the right of the line x œ b and rises to the left. If a œ 0, the graph reduces to the horizontal line y œ c. As kak increases, the slope at any given point x œ x! increases in magnitude and the graph becomes steeper. As kak decreases, the slope at x! decreases in magnitude and the graph rises or falls more gradually. (b) Increasing b shifts the graph to the left; decreasing b shifts it to the right. (c) Increasing c shifts the graph upward; decreasing c shifts it downward. 27. If m 0, the x-intercept of y œ mx 2 must be negative. If m 0, then the x-intercept exceeds Ê 0 œ mx 2 and x
" #
Ê x œ m2
" #
Ê 0 m 4.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
" #
Chapter 1 Additional and Advanced Exercises 28. Each of the triangles pictured has the same base b œ v?t œ v(1 sec). Moreover, the height of each triangle is the same value h. Thus "# (base)(height) œ
" #
bh
œ A" œ A# œ A$ œ á . In conclusion, the object sweeps out equal areas in each one second interval.
29. (a) By Exercise #95 of Section 1.2, the coordinates of P are ˆ a# 0 ß b# 0 ‰ œ ˆ #a ß b# ‰ . Thus the slope of OP œ
(b)
?y ?x
œ
b/2 a/2
œ
b a . b 0 The slope of AB œ 0a œ ba . The line # of their slopes is " œ ˆ ba ‰ ˆ ba ‰ œ ba#
segments AB and OP are perpendicular when the product . Thus, b# œ a# Ê a œ b (since both are positive). Therefore, AB
is perpendicular to OP when a œ b.
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65
66
Chapter 1 Preliminaries
NOTES:
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
CHAPTER 2 LIMITS AND CONTINUITY 2.1 RATES OF CHANGE AND LIMITS 1. (a) Does not exist. As x approaches 1 from the right, g(x) approaches 0. As x approaches 1 from the left, g(x) approaches 1. There is no single number L that all the values g(x) get arbitrarily close to as x Ä 1. (b) 1 (c) 0 2. (a) 0 (b) 1 (c) Does not exist. As t approaches 0 from the left, f(t) approaches 1. As t approaches 0 from the right, f(t) approaches 1. There is no single number L that f(t) gets arbitrarily close to as t Ä 0. 3. (a) True (d) False
(b) True (e) False
(c) False (f) True
4. (a) False (d) True
(b) False (e) True
(c) True
5.
x
lim x Ä 0 kx k x kx k
does not exist because
x kx k
œ
x x
œ 1 if x 0 and
approaches 1. As x approaches 0 from the right,
x kx k
x kxk
œ
x x
œ 1 if x 0. As x approaches 0 from the left,
approaches 1. There is no single number L that all
the function values get arbitrarily close to as x Ä 0. 6. As x approaches 1 from the left, the values of
" x 1
become increasingly large and negative. As x approaches 1
from the right, the values become increasingly large and positive. There is no one number L that all the function values get arbitrarily close to as x Ä 1, so lim x" 1 does not exist. xÄ1
7. Nothing can be said about f(x) because the existence of a limit as x Ä x! does not depend on how the function is defined at x! . In order for a limit to exist, f(x) must be arbitrarily close to a single real number L when x is close enough to x! . That is, the existence of a limit depends on the values of f(x) for x near x! , not on the definition of f(x) at x! itself. 8. Nothing can be said. In order for lim f(x) to exist, f(x) must close to a single value for x near 0 regardless of xÄ0
the value f(0) itself. 9. No, the definition does not require that f be defined at x œ 1 in order for a limiting value to exist there. If f(1) is defined, it can be any real number, so we can conclude nothing about f(1) from lim f(x) œ 5. xÄ1
10. No, because the existence of a limit depends on the values of f(x) when x is near 1, not on f(1) itself. If lim f(x) exists, its value may be some number other than f(1) œ 5. We can conclude nothing about lim f(x), xÄ1
xÄ1
whether it exists or what its value is if it does exist, from knowing the value of f(1) alone.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Chapter 2 Limits and Continuity
11. (a) f(x) œ ax# *b/(x 3) x 3.1 f(x) 6.1 2.9 5.9
x f(x)
3.01 6.01
3.001 6.001
3.0001 6.0001
3.00001 6.00001
3.000001 6.000001
2.99 5.99
2.999 5.999
2.9999 5.9999
2.99999 5.99999
2.999999 5.999999
The estimate is lim f(x) œ 6. x Ä $
(b)
(c) f(x) œ
x# 9 x3
œ
(x 3)(x 3) x3
œ x 3 if x Á 3, and lim (x 3) œ 3 3 œ 6. x Ä $
12. (a) g(x) œ ax# #b/ Šx È2‹ x g(x)
1.4 2.81421
1.41 2.82421
1.414 2.82821
1.4142 2.828413
1.41421 2.828423
1.414213 2.828426
(b)
(c) g(x) œ
x# 2 x È2
œ
Šx È2‹ Šx È2‹ Šx È2‹
œ x È2 if x Á È2, and
13. (a) G(x) œ (x 6)/ ax# 4x 12b x 5.9 5.99 G(x) .126582 .1251564 x G(x)
6.1 .123456
6.01 .124843
5.999 .1250156 6.001 .124984
lim
x Ä È#
5.9999 .1250015 6.0001 .124998
Šx È2‹ œ È2 È2 œ 2È2.
5.99999 .1250001 6.00001 .124999
5.999999 .1250000 6.000001 .124999
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 2.1 Rates of Change and Limits (b)
(c) G(x) œ
x6 ax# 4x 12b
œ
x6 (x 6)(x 2)
œ
" x#
14. (a) h(x) œ ax# 2x 3b / ax# 4x 3b x 2.9 2.99 h(x) 2.052631 2.005025 x h(x)
3.1 1.952380
3.01 1.995024
"
if x Á 6, and lim
x Ä ' x 2
œ
" ' 2
œ "8 œ 0.125.
2.999 2.000500
2.9999 2.000050
2.99999 2.000005
2.999999 2.0000005
3.001 1.999500
3.0001 1.999950
3.00001 1.999995
3.000001 1.999999
(b)
(c) h(x) œ
x# 2x 3 x# 4x 3
œ
(x 3)(x 1) (x 3)(x 1)
œ
x1 x1
15. (a) f(x) œ ax# 1b / akxk 1b x 1.1 1.01 f(x) 2.1 2.01 x f(x)
.9 1.9
.99 1.99
if x Á 3, and lim
x1
x Ä $ x1
œ
31 31
œ
4 #
œ 2.
1.001 2.001
1.0001 2.0001
1.00001 2.00001
1.000001 2.000001
.999 1.999
.9999 1.9999
.99999 1.99999
.999999 1.999999
(b)
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69
70
Chapter 2 Limits and Continuity (c) f(x) œ
x# " kx k 1
(x 1)(x 1)
1 œ (x x1)(x 1) (x 1)
œ x 1, x 0 and x Á 1 , and lim (1 x) œ 1 (1) œ 2. x Ä 1 œ 1 x, x 0 and x Á 1
16. (a) F(x) œ ax# 3x 2b / a2 kxkb x 2.1 2.01 F(x) 1.1 1.01 1.9 .9
x F(x)
1.99 .99
2.001 1.001
2.0001 1.0001
2.00001 1.00001
2.000001 1.000001
1.999 .999
1.9999 .9999
1.99999 .99999
1.999999 .999999
(b)
(c) F(x) œ
x# 3x 2 2 kx k
(x 2)(x 1)
œ (x 2)(x# x") 2x
17. (a) g()) œ (sin ))/) ) .1 g()) .998334
, x 0 , and lim (x 1) œ 2 1 œ 1. x Ä # œ x 1, x 0 and x Á 2
.01 .999983
.001 .999999
.0001 .999999
.00001 .999999
.000001 .999999
.1 .998334
.01 .999983
.001 .999999
.0001 .999999
.00001 .999999
.000001 .999999
18. (a) G(t) œ (1 cos t)/t# t .1 G(t) .499583
.01 .499995
.001 .499999
.0001 .5
.00001 .5
.000001 .5
.1 .499583
.01 .499995
.001 .499999
.0001 .5
.00001 .5
.000001 .5
) g()) lim g()) œ 1
)Ä!
(b)
t G(t)
lim G(t) œ 0.5
tÄ!
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 2.1 Rates of Change and Limits (b)
Graph is NOT TO SCALE 19. (a) f(x) œ x"ÎÐ"xÑ x .9 f(x) .348678 x f(x)
1.1 .385543
.99 .366032
.999 .367695
.9999 .367861
.99999 .367877
.999999 .367879
1.01 .369711
1.001 .368063
1.0001 .367897
1.00001 .367881
1.000001 .367878
lim f(x) ¸ 0.36788
xÄ1
(b)
Graph is NOT TO SCALE. Also the intersection of the axes is not the origin: the axes intersect at the point (1ß 2.71820). 20. (a) f(x) œ a3x 1b /x x .1 f(x) 1.161231
.01 1.104669
.001 1.099215
.0001 1.098672
.00001 1.098618
.000001 1.098612
.1 1.040415
.01 1.092599
.001 1.098009
.0001 1.098551
.00001 1.098606
.000001 1.098611
x f(x)
lim f(x) ¸ 1.0986
xÄ!
(b)
21. lim 2x œ 2(2) œ 4
22. lim 2x œ 2(0) œ 0
23. lim" (3x 1) œ 3 ˆ "3 ‰ 1 œ 0
24. lim
xÄ#
xÄ
$
xÄ!
1
x Ä 1 3x1
œ
" 3(1) 1
œ #"
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
71
72 25.
Chapter 2 Limits and Continuity lim 3x(2x 1) œ 3(1)(2(1) 1) œ 9
26.
x Ä "
1 #
27. lim1 x sin x œ xÄ
#
1 #
sin
œ
1 #
28. xlim Ä1
29. (a)
?f ?x
œ
f(3) f(2) 3#
30. (a)
?g ?x
œ
g(1) g(1) 1 (1)
31. (a)
?h ?t
œ
h ˆ 341 ‰ h ˆ 14 ‰ 1 31 4 4
œ
?g ?t
œ
g(1) g(0) 10
(2 1) (2 1) 10
32. (a) 33.
?R ?)
œ
R(2) R(0) 20
34.
?P ?)
œ
P(2) P(1) 21
35. (a)
œ œ
28 9 1
œ
œ
œ
1 1 2
Q% (18ß 550)
œ
3 3
œ 1
" 1 1
œ
" 1 1
œ
f(1) f(") 1 (1)
œ
20 #
œ1
œ0
(b)
?g ?x
œ
g(0)g(2) 0(2)
œ
04 #
œ 2
(b)
?h ?t
œ
h ˆ 1# ‰ h ˆ 16 ‰ 11 # 6
œ
?g ?t
œ
g(1) g(1) 1 (1)
œ
1 1 1 #
œ 14
œ
650 225 20 10 650 375 20 14 650 475 20 16.5 650 550 20 18
Q$ (16.5ß 475)
cos 1 1 1
œ
?f ?x
œ 12
3" #
(b)
0 È3 1 3
œ
3 È 3 1
(2 1) (2 ") #1
œ0
œ1 œ22œ0
Slope of PQ œ
Q# (14ß 375)
œ
3(1)# 2(1)1
(b)
(8 16 10)(" % &) 1
Q" (10ß 225)
cos x 1 1
œ
œ 19
È 8 1 È 1 #
Q
3x#
lim x Ä 1 2x1
?p ?t
œ 42.5 m/sec œ 45.83 m/sec œ 50.00 m/sec œ 50.00 m/sec
(b) At t œ 20, the Cobra was traveling approximately 50 m/sec or 180 km/h. 36. (a)
Slope of PQ œ
Q Q" (5ß 20) Q# (7ß 39) Q$ (8.5ß 58) Q% (9.5ß 72)
80 20 10 5 80 39 10 7 80 58 10 8.5 80 72 10 9.5
?p ?t
œ 12 m/sec œ 13.7 m/sec œ 14.7 m/sec œ 16 m/sec
(b) Approximately 16 m/sec 37. (a)
(b)
?p ?t
œ
174 62 1994 1992
œ
112 #
œ 56 thousand dollars per year
(c) The average rate of change from 1991 to 1992 is ??pt œ The average rate of change from 1992 to 1993
is ??pt
œ
62 27 1992 1991 111 62 1993 1992
œ 35 thousand dollars per year. œ 49 thousand dollars per year.
So, the rate at which profits were changing in 1992 is approximatley "# a35 49b œ 42 thousand dollars per year.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 2.1 Rates of Change and Limits 38. (a) F(x) œ (x 2)/(x 2) x 1.2 F(x) 4.0 ?F ?x ?F ?x ?F ?x
œ
?g ?x ?g ?x
œ
œ œ
1.1 3.4
1.01 3.04
1.001 3.004
1.0001 3.0004
1 3
4.0 (3) œ 5.0; 1.2 1 3.04 (3) œ 4.04; 1.01 1 3.!!!% (3) œ 4.!!!%; 1.0001 1
?F ?x ?F ?x
œ œ
3.4 (3) œ 4.4; 1.1 1 3.004 (3) œ 4.!!%; 1.001 1
È g(2) g(1) œ #21" ¸ 0.414213 21 È1 h" g(1 h) g(1) (1 h) 1 œ h
?g ?x
œ
g(1.5) g(1) 1.5 1
(b) The rate of change of F(x) at x œ 1 is 4. 39. (a)
œ
œ
È1.5 " 0.5
¸ 0.449489
(b) g(x) œ Èx 1h È1 h
1.1 1.04880
1.01 1.004987
1.001 1.0004998
1.0001 1.0000499
1.00001 1.000005
1.000001 1.0000005
ŠÈ1 h 1‹ /h
0.4880
0.4987
0.4998
0.499
0.5
0.5
(c) The rate of change of g(x) at x œ 1 is 0.5. (d) The calculator gives lim hÄ! 40. (a) i) ii) (b)
f(3) f(2) 32 f(T) f(2) T#
œ œ
"" 3 #
1 " " T # T#
T f(T) af(T) f(2)b/aT 2b
œ œ
" 6
1
È1 h" h
œ 6"
#TT T#
2 #T
œ "# .
œ
2.1 0.476190 0.2381
2T #T(T 2)
œ
2T #T(2 T)
2.01 0.497512 0.2488
œ #"T , T Á 2
2.001 0.499750 0.2500
2.0001 0.4999750 0.2500
2.00001 0.499997 0.2500
2.000001 0.499999 0.2500
(c) The table indicates the rate of change is 0.25 at t œ 2. " ‰ (d) lim ˆ #T œ 4" TÄ#
41-46. Example CAS commands: Maple: f := x -> (x^4 16)/(x 2); x0 := 2; plot( f(x), x œ x0-1..x0+1, color œ black, title œ "Section 2.1, #41(a)" ); limit( f(x), x œ x0 ); In Exercise 43, note that the standard cube root, x^(1/3), is not defined for x (surd(x+1, 3) 1)/x. Mathematica: (assigned function and values for x0 and h may vary) Clear[f, x] f[x_]:=(x3 x2 5x 3)/(x 1)2 x0= 1; h = 0.1; Plot[f[x],{x, x0 h, x0 h}] Limit[f[x], x Ä x0]
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
73
74
Chapter 2 Limits and Continuity
2.2 CALCULATING LIMITS USING THE LIMIT LAWS 1. 3. 4. 5.
lim (2x 5) œ 2(7) 5 œ 14 5 œ 9
lim ax# 5x 2b œ (2)# 5(2) 2 œ 4 10 2 œ 4 lim ax$ 2x# 4x 8b œ (2)$ 2(2)# 4(2) 8 œ 8 8 8 8 œ 16
x Ä #
lim 8(t 5)(t 7) œ 8(6 5)(6 7) œ 8 x3
œ
9.
lim y Ä & 5 y
y#
23 26
# y Ä # y 5y 6
œ
5 8
(5)# 5 (5)
œ
y2
10. lim
13.
6.
tÄ'
lim x Ä # x6
12.
lim (10 3x) œ 10 3(12) œ 10 36 œ 26
x Ä 1#
xÄ#
7.
11.
2.
x Ä (
œ
8. œ
25 10
œ
22 (2)# 5(#) 6
lim# 3s(2s 1) œ 3 ˆ 23 ‰ 2 ˆ 23 ‰ 1‘ œ 2 ˆ 43 1‰ œ
sÄ
$
4
lim x Ä & x7
œ
4 57
œ
4 #
œ 2
5 #
œ
4 4 10 6
œ
4 #0
œ
" 5
lim 3(2x 1)# œ 3(2(1) 1)# œ 3(3)# œ 27
x Ä "
lim (x 3)"*)% œ (4 3)"*)% œ (1)"*)% œ 1
x Ä %
%
lim (5 y)%Î$ œ [5 (3)]%Î$ œ (8)%Î$ œ ˆ(8)"Î$ ‰ œ 2% œ 16
y Ä $
14. lim (2z 8)"Î$ œ (2(0) 8)"Î$ œ (8)"Î$ œ 2 zÄ!
15. lim
3
œ
3 È3(0) 1 1
œ
3 È1 1
œ
3 2
16. lim
5
œ
5 È5(0) 4 2
œ
5 È4 #
œ
5 4
17. lim
È3h 1 " h
h Ä ! È3h 1 1 h Ä ! È5h 4 2
hÄ0
œ
3 È" "
œ
È5h 4 2 h hÄ0 5 È4 2
19. lim
œ
x5
# x Ä & x 25
20. 21.
œ lim
a3h "b 1
È5h 4 2 h hÄ0
†
È5h 4 2 È5h 4 2
œ lim
a5h 4b 4
h Ä 0 hŠÈ3h 1 "‹
œ lim
3h
œ lim
5h
h Ä 0 hŠÈ3h 1 "‹
œ lim
3
h Ä 0 È3h1"
h Ä 0 hŠÈ5h 4 2‹
h Ä 0 hŠÈ5h 4 2‹
œ lim
5 4 x5
œ lim
x Ä & (x 5)(x 5)
x3
lim
È3h 1 1 È3h 1 1
œ lim
lim # x Ä $ x 4x 3 x Ä &
†
hÄ0
$ #
18. lim œ
È3h 1 " h
œ lim
x# 3x "0 x5
œ lim
œ lim
x3
x Ä $ (x 3)(x 1)
œ lim
x Ä &
1
x Ä & x5
(x 5)(x 2) x5
œ
œ lim
" 55
œ
" 10
1
œ
" 3 1
x Ä $ x 1
œ "2
œ lim (x 2) œ & # œ 7 x Ä &
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
5
h Ä 0 È5h 4 2
2 3
Section 2.2 Calculating Limits Using the Limit Laws (x 5)(x 2) x2
22. lim
x# 7x "0 x#
œ lim
23. lim
t# t 2 t# 1
t Ä " (t 1)(t 1)
xÄ#
tÄ"
t# 3t 2
lim # t Ä " t t 2
25.
lim $ # x Ä # x 2x
2x 4
5y$ 8y#
u% "
$ u Ä 1 u 1
œ lim
y# (5y 8)
œ lim
œ lim
4x x#
œ lim
x Ä 1 Èx 3 2
lim
x Ä "
œ lim
xÄ%
œ
x Ä "
35.
x2 x Ä 2 È x # 5 3
œ lim
œ
lim
"
œ lim
x ˆ2 È x ‰ ˆ 2 È x ‰ 2 Èx
œ lim
xÄ1
x1
œ lim
x2
x Ä 2 Èx# 12 4
œ lim
x Ä 2
(x 3) Š2 Èx# 5‹
9 x#
œ
lim
12 32
œ
3 8
" 6
œ lim x ˆ2 Èx‰ œ 4(2 2) œ 16 xÄ%
(x 1) ˆÈx 3 #‰ (x 3) 4
2 33
œ lim ŠÈx 3 #‹ xÄ1
œ "3
ax# 12b 16 x Ä 2 (x 2) ŠÈx# 12 4‹
œ lim 4 È16 4
œ
œ lim
" 2
ax 2b ŠÈx# 5 3‹ ax # 5 b 9
x Ä 2
œ
Š2 Èx# 5‹ Š2 Èx# 5‹
x Ä 3
x Ä 3 (x 3) Š2 Èx# 5‹
È x# 5 3 x2
œ
œ
4 3
ax # 8 b * x Ä 1 (x 1) ŠÈx# 8 $‹
ax 2b ŠÈx# 5 3‹
œ lim
444 (4)(8)
œ
œ
œ lim
œ
œ
(1 1)(1 1) 111
" È9 3
œ
x Ä * Èx 3
x Ä 2 ŠÈx# 5 3‹ ŠÈx# 5 3‹
(x 2)(x 2)
œ
v# 2v 4 (v 2) av# 4b vÄ#
(x 2) ŠÈx# 12 4‹
œ lim
œ #"
œ lim
ŠÈx# 12 4‹ ŠÈx# 12 4‹
xÄ2
œ 13
au# "b (u 1) u# u 1
x Ä 1 È x # ) $
ax 2b ŠÈx# 5 3‹
2 È x# 5 x3 x Ä 3
lim
uÄ1
œ lim
(x 2)(x 2)
lim
x Ä 2
8 16
(x 1) ŠÈx# 8 $‹
x Ä 2 (x 2) ŠÈx# 12 4‹
lim
œ
ŠÈx# 8 $‹ ŠÈx# 8 $‹
lim
(x 1)(x 1)
Èx# 12 4 x2 xÄ2
œ
5y 8
œ È4 2 œ 4
33. lim
34.
œ 21
x Ä 1 ˆÈ x 3 # ‰ ˆ È x 3 # ‰
x Ä 1 (x 1) ŠÈx# ) $‹
œ lim
2 4
(x 1) ˆÈx 3 2‰
œ lim
È x# 8 3 x1
œ lim
x(4 x)
x Ä % 2 Èx
x1
31. lim
Èx 3
x Ä * ˆÈ x 3 ‰ ˆ È x 3 ‰
x Ä % 2 Èx
œ
2
œ lim
(v 2) av# 2v 4b (v 2)(v 2) av# 4b vÄ#
3 #
1 2 1 2
# y Ä ! 3y 16
au# "b (u 1)(u 1) au# u 1b (u 1)
œ
œ
# x Ä # x
# # y Ä ! y a3y 16b
Èx 3 x9
t2
t Ä " t 2
œ lim
uÄ1
12 11
œ
œ lim
2(x 2)
œ lim
30. lim
t2
œ lim
v$ 8 % 16 v vÄ#
xÄ*
32.
t Ä " (t 2)(t 1)
œ lim
28. lim
29. lim
xÄ#
t Ä " t1
(t 2)(t 1)
œ lim
œ lim (x 5) œ 2 5 œ 3
œ lim
# x Ä # x (x 2)
% # y Ä 0 3y 16y
27. lim
(t 2)(t 1)
œ lim
24.
26. lim
xÄ#
È9 3 4
œ 23
œ lim
(3 x)(3 x)
4 ax # 5 b
x Ä 3 (x 3) Š2 Èx# 5‹
x Ä 3 (x 3) Š2 Èx# 5‹
œ lim
3x
x Ä 3 2 È x # 5
œ
6 2 È4
œ
3 2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
75
76
Chapter 2 Limits and Continuity 4x x Ä 4 5 È x# 9
œ lim
a4 xb Š5 Èx# 9‹
œ lim
36. lim
x Ä 4 Š5 Èx# 9‹ Š5 Èx# 9‹
a4 xb Š5 Èx# 9‹
xÄ4
16 x#
œ lim
(4 x)(4 x)
xÄ4
25 ax# 9b
xÄ4
a4 xb Š5 Èx# 9‹
œ lim
a4 xb Š5 Èx# 9‹
5 È x# 9 4x
œ lim
xÄ4
œ
5 È25 8
œ
5 4
37. (a) quotient rule (b) difference and power rules (c) sum and constant multiple rules 38. (a) quotient rule (b) power and product rules (c) difference and constant multiple rules 39. (a) xlim f(x) g(x) œ ’xlim f(x)“ ’ x lim g(x)“ œ (5)(2) œ 10 Äc Äc Äc (b) xlim 2f(x) g(x) œ 2 ’xlim f(x)“ ’ xlim g(x)“ œ 2(5)(2) œ 20 Äc Äc Äc (c) xlim [f(x) 3g(x)] œ xlim f(x) 3 xlim g(x) œ 5 3(2) œ 1 Äc Äc Äc lim f(x) f(x) 5 5 xÄc (d) xlim œ lim f(x) lim g(x) œ 5(2) œ 7 Ä c f(x) g(x) x
40. (a) (b) (c) (d) 41. (a) (b) (c) (d) 42. (a) (b) (c)
Äc
Äc
lim [g(x) 3] œ lim g(x) lim 3 œ $ $ œ !
xÄ%
xÄ%
xÄ%
lim xf(x) œ lim x † lim f(x) œ (4)(0) œ 0
xÄ%
xÄ%
#
xÄ%
#
lim [g(x)] œ ’ lim g(x)“ œ [3]# œ 9
xÄ%
g(x) x Ä % f(x) 1
lim
xÄ%
œ
Ä%
lim g(x)
x
lim f(x) lim 1
xÄ%
xÄ%
œ
3 01
œ3
lim [f(x) g(x)] œ lim f(x) lim g(x) œ 7 (3) œ 4
xÄb
xÄb
xÄb
lim f(x) † g(x) œ ’ lim f(x)“ ’ lim g(x)“ œ (7)(3) œ 21
xÄb
xÄb
xÄb
lim 4g(x) œ ’ lim 4“ ’ lim g(x)“ œ (4)(3) œ 12
xÄb
xÄb
xÄb
lim f(x)/g(x) œ lim f(x)/ lim g(x) œ
xÄb
xÄb
xÄb
7 3
œ 73
lim [p(x) r(x) s(x)] œ lim p(x) lim r(x) lim s(x) œ 4 0 (3) œ 1
x Ä #
x Ä #
x Ä #
x Ä #
lim p(x) † r(x) † s(x) œ ’ lim p(x)“ ’ lim r(x)“ ’ lim s(x)“ œ (4)(0)(3) œ 0
x Ä #
x Ä #
(1 h)# 1# h hÄ!
œ lim
hÄ!
(2 h)# (2)# h
45. lim
[3(2 h) 4] [3(2) 4] h
hÄ!
x Ä #
x Ä #
44. lim
hÄ!
x Ä #
lim [4p(x) 5r(x)]/s(x) œ ’4 lim p(x) 5 lim r(x)“ ‚ lim s(x) œ [4(4) 5(0)]/3 œ
x Ä #
43. lim
"‰ ˆ #" h ‰ ˆ # h hÄ!
46. lim
x
1 2h h# 1 h
œ lim
hÄ!
œ lim
hÄ!
44hh# 4 h
œ lim
hÄ!
œ lim
3h
hÄ! h
2 2 h " 2h
œ lim
x Ä #
h(2 h) h
hÄ!
x Ä #
œ lim (2 h) œ 2
h(h 4) h
hÄ!
œ lim (h 4) œ 4 hÄ!
œ3
œ lim
hÄ!
2 (2 h) 2h(# h)
œ lim
h
h Ä ! h(4 2h)
œ "4
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
"6 3
Section 2.2 Calculating Limits Using the Limit Laws È7 h È7 h hÄ!
47. lim
œ lim
ŠÈ7 h È7‹ ŠÈ7 h È7‹
œ lim
h ŠÈ7 h È7‹
hÄ!
h
h Ä ! h ŠÈ7hÈ7‹
È3(0 h) 1 È3(0) 1 h hÄ! 3h
h Ä ! h ŠÈ3h 1 "‹
œ
h Ä ! È 7 h È 7
48. lim
œ lim
"
œ lim
œ lim
" #È 7
ŠÈ3h 1 "‹ ŠÈ3h 1 "‹
œ lim
h ŠÈ3h 1 "‹
hÄ!
œ lim
œ
3
h Ä ! È3h 1 1
(7 h) 7
h Ä ! h ŠÈ7 h È7‹
(3h 1) "
œ lim
h Ä ! h ŠÈ3h 1 1 ‹
3 #
49. lim È5 2x# œ È5 2(0)# œ È5 and lim È5 x# œ È5 (0)# œ È5; by the sandwich theorem, xÄ!
xÄ!
lim f(x) œ È5
xÄ!
50. lim a2 x# b œ 2 0 œ 2 and lim 2 cos x œ 2(1) œ 2; by the sandwich theorem, lim g(x) œ 2 xÄ!
51. (a)
xÄ!
lim Š1
xÄ!
x# 6‹
œ1
0 6
xÄ!
œ 1 and lim 1 œ 1; by the sandwich theorem, lim
(b) For x Á 0, y œ (x sin x)/(2 2 cos x) lies between the other two graphs in the figure, and the graphs converge as x Ä 0.
52. (a)
lim Š "#
xÄ!
lim
xÄ!
1cos x x#
x# 24 ‹
œ lim
1
xÄ! #
lim
x#
x Ä ! #4
œ
" #
x sin x
x Ä ! 22 cos x
xÄ!
0œ
" #
and lim
"
xÄ! #
œ1
œ "# ; by the sandwich theorem,
œ "# .
(b) For all x Á 0, the graph of f(x) œ (1 cos x)/x# lies between the line y œ "# and the parabola yœ
" #
x# /24, and the graphs converge as x Ä 0.
53. xlim f(x) exists at those points c where xlim x% œ xlim x# . Thus, c% œ c# Ê c# a1 c# b œ 0 Äc Äc Äc Ê c œ 0, 1, or 1. Moreover, lim f(x) œ lim x# œ 0 and lim f(x) œ lim f(x) œ 1. xÄ!
xÄ!
x Ä 1
xÄ1
54. Nothing can be concluded about the values of f, g, and h at x œ 2. Yes, f(2) could be 0. Since the conditions of the sandwich theorem are satisfied, lim f(x) œ 5 Á 0. xÄ#
55. 1 œ lim
xÄ%
f(x)5 x 2
lim f(x) lim 5 % xÄ% œ xÄlim x lim 2 œ xÄ%
xÄ%
lim f(x) 5
xÄ%
%#
Ê lim f(x) 5 œ 2(1) Ê lim f(x) œ 2 5 œ 7. xÄ%
xÄ%
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
77
78
Chapter 2 Limits and Continuity
56. (a) 1 œ lim
f(x) x#
lim f(x) lim f(x) # xÄ# œ xÄlim Ê % x# œ
(b) 1 œ lim
f(x) x#
œ ’ lim
x Ä # x Ä #
xÄ#
x Ä #
57. (a) 0 œ 3 † 0 œ ’ lim
xÄ#
lim f(x) œ 4.
x Ä #
f(x) lim x" “ x “ ’x Ä #
œ ’ lim
x Ä #
f(x) ˆ " ‰ x “ #
Ê
lim
x Ä #
f(x) x
œ 2.
f(x) 5 x # “ ’xlim Ä#
5 (x 2)“ œ lim ’Š f(x) x # ‹ (x 2)“ œ lim [f(x) 5] œ lim f(x) 5
f(x) 5 x # “ ’xlim Ä#
(x 2)“ Ê lim f(x) œ 5 as in part (a).
xÄ#
Ê lim f(x) œ 5.
xÄ#
xÄ#
xÄ#
(b) 0 œ 4 † 0 œ ’ lim
xÄ#
58. (a) 0 œ 1 † 0 œ ’ lim
f(x) # “ ’ lim xÄ! x xÄ!
(b) 0 œ 1 † 0 œ 59. (a)
lim x sin
xÄ!
(b) 1 Ÿ sin
60. (a)
" x
’ lim f(x) # “ ’ lim xÄ! x xÄ!
" x
xÄ#
#
x“ œ ’ lim
f(x)
# xÄ! x
x“ œ
lim ’ f(x) x# xÄ!
# “ ’ lim x# “ œ lim ’ f(x) x# † x “ œ lim f(x). That is, lim f(x) œ 0.
xÄ!
† x“ œ
xÄ!
lim f(x) . xÄ! x
That is,
xÄ!
lim f(x) xÄ! x
œ 0.
œ0
Ÿ 1 for x Á 0:
x 0 Ê x Ÿ x sin
" x
Ÿ x Ê lim x sin
" x
œ 0 by the sandwich theorem;
x 0 Ê x x sin
" x
x Ê lim x sin
" x
œ 0 by the sandwich theorem.
xÄ! xÄ!
lim x# cos ˆ x"$ ‰ œ 0
xÄ!
(b) 1 Ÿ cos ˆ x"$ ‰ Ÿ 1 for x Á 0 Ê x# Ÿ x# cos ˆ x"$ ‰ Ÿ x# Ê lim x# cos ˆ x"$ ‰ œ 0 by the sandwich theorem since lim x# œ 0.
xÄ!
xÄ!
2.3 PRECISE DEFINITION OF A LIMIT 1. Step 1: Step 2:
kx 5k $ Ê $ x 5 $ Ê $ 5 x $ 5 $ 5 œ 7 Ê $ œ 2, or $ 5 œ 1 Ê $ œ 4. The value of $ which assures kx 5k $ Ê 1 x 7 is the smaller value, $ œ 2.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
xÄ!
Section 2.3 Precise Definition of a Limit 2. Step 1: Step 2:
kx 2k $ Ê $ x 2 $ Ê $ # x $ 2 $ 2 œ 1 Ê $ œ 1, or $ 2 œ 7 Ê $ œ 5. The value of $ which assures kx 2k $ Ê 1 x 7 is the smaller value, $ œ 1.
Step 1: Step 2:
kx (3)k $ Ê $ x $ $ Ê $ 3 x $ 3 $ 3 œ 7# Ê $ œ "# , or $ $ œ "# Ê $ œ 5# .
3.
The value of $ which assures kx (3)k $ Ê 7# x "# is the smaller value, $ œ "# .
4.
Step 1:
¸x ˆ 3# ‰¸ $ Ê $ x
Step 2:
$
Step 1:
¸x "# ¸ $ Ê $ x
Step 2:
$
œ
3 #
$ Ê $ " #
3 #
x$
3 #
Ê $ œ #, or $ œ Ê $ œ 1. The value of $ which assures ¸x ˆ 3# ‰¸ $ Ê 7# x "# is the smaller value, $ œ ". 3 #
7 #
3 #
5. " #
$ Ê $
" #
x$
" or $ #" œ 47 Ê $ œ 14 . " 4 The value of $ which assures ¸x # ¸ $ Ê 9 x
œ
4 9
Ê $œ
" 18 ,
" #
4 7
" #
is the smaller value, $ œ
" 18 .
6.
Step 1: Step 2:
kx 3k $ Ê $ x 3 $ Ê $ 3 x $ 3 $ $ œ 2.7591 Ê $ œ 0.2409, or $ $ œ 3.2391 Ê $ œ 0.2391. The value of $ which assures kx 3k $ Ê 2.7591 x 3.2391 is the smaller value, $ œ 0.2391.
7. Step 1: Step 2:
kx 5k $ Ê $ x 5 $ Ê $ 5 x $ 5 From the graph, $ 5 œ 4.9 Ê $ œ 0.1, or $ 5 œ 5.1 Ê $ œ 0.1; thus $ œ 0.1 in either case.
8. Step 1: Step 2:
kx (3)k $ Ê $ x 3 $ Ê $ 3 x $ 3 From the graph, $ 3 œ 3.1 Ê $ œ 0.1, or $ 3 œ 2.9 Ê $ œ 0.1; thus $ œ 0.1.
9. Step 1: Step 2:
kx 1k $ Ê $ x 1 $ Ê $ 1 x $ 1 9 7 From the graph, $ 1 œ 16 Ê $ œ 16 , or $ 1 œ 25 16 Ê $ œ
10. Step 1: Step 2:
kx 3k $ Ê $ x 3 $ Ê $ 3 x $ 3 From the graph, $ 3 œ 2.61 Ê $ œ 0.39, or $ 3 œ 3.41 Ê $ œ 0.41; thus $ œ 0.39.
11. Step 1:
kx 2k $ Ê $ x 2 $ Ê $ 2 x $ 2 From the graph, $ 2 œ È3 Ê $ œ 2 È3 ¸ 0.2679, or $ 2 œ È5 Ê $ œ È5 2 ¸ 0.2361; thus $ œ È5 2.
Step 2:
9 16 ;
thus $ œ
7 16 .
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
79
80
Chapter 2 Limits and Continuity
12. Step 1: Step 2:
kx (1)k $ Ê $ x 1 $ Ê $ 1 x $ 1 From the graph, $ 1 œ thus $ œ
È5 2 # .
È5 #
Ê $œ
È5 2 #
¸ 0.1180, or $ 1 œ
13. Step 1: Step 2:
kx (1)k $ Ê $ x 1 $ Ê $ 1 x $ 1 7 16 From the graph, $ 1 œ 16 9 Ê $ œ 9 ¸ 0.77, or $ 1 œ 25 Ê
14. Step 1:
¸x "# ¸ $ Ê $ x
Step 2:
From the graph, $ thus $ œ 0.00248.
" #
œ
" # 1 2.01
$ Ê $ Ê $œ
1 2
" #
x$
" #.01
" #
¸ 0.00248, or $
" #
œ
È3 #
9 25
Ê $œ
2 È3 #
œ 0.36; thus $ œ
1 1.99
Ê $œ
1 1.99
¸ 0.1340;
9 25
" #
œ 0.36.
¸ 0.00251;
15. Step 1: Step 2:
k(x 1) 5k 0.01 Ê kx 4k 0.01 Ê 0.01 x 4 0.01 Ê 3.99 x 4.01 kx 4k $ Ê $ x 4 $ Ê $ 4 x $ 4 Ê $ œ 0.01.
16. Step 1:
k(2x 2) (6)k 0.02 Ê k2x 4k 0.02 Ê 0.02 2x 4 0.02 Ê 4.02 2x 3.98 Ê 2.01 x 1.99 kx (2)k $ Ê $ x 2 $ Ê $ 2 x $ 2 Ê $ œ 0.01.
Step 2: 17. Step 1: Step 2: 18. Step 1: Step 2:
¹Èx 1 "¹ 0.1 Ê 0.1 Èx 1 " 0.1 Ê 0.9 Èx 1 1.1 Ê 0.81 x 1 1.21 Ê 0.19 x 0.21 kx 0k $ Ê $ x $ . Then, $ œ !Þ"* Ê $ œ !Þ"* or $ œ !Þ#"; thus, $ œ 0.19. ¸Èx "# ¸ 0.1 Ê 0.1 Èx "# 0.1 Ê 0.4 Èx 0.6 Ê 0.16 x 0.36 ¸x "4 ¸ $ Ê $ x 4" $ Ê $ 4" B $ 4" . Then, $
19. Step 1: Step 2:
20. Step 1: Step 2:
21. Step 1: Step 2:
22. Step 1: Step 2:
" 4
œ 0.16 Ê $ œ 0.09 or $
" 4
œ 0.36 Ê $ œ 0.11; thus $ œ 0.09.
¹È19 x $¹ " Ê " È19 x $ 1 Ê 2 È19 x % Ê 4 19 x 16 Ê % x 19 16 Ê 15 x 3 or 3 x 15 kx 10k $ Ê $ x 10 $ Ê $ 10 x $ 10. Then $ 10 œ 3 Ê $ œ 7, or $ 10 œ 15 Ê $ œ 5; thus $ œ 5. ¹Èx 7 4¹ 1 Ê " Èx 7 % 1 Ê 3 Èx 7 5 Ê 9 x 7 25 Ê 16 x 32 kx 23k $ Ê $ x 23 $ Ê $ 23 x $ 23. Then $ 23 œ 16 Ê $ œ 7, or $ 23 œ 32 Ê $ œ 9; thus $ œ 7. ¸ "x 4" ¸ 0.05 Ê 0.05
" x
" 4
0.05 Ê 0.2
" x
0.3 Ê
kx 4k $ Ê $ x 4 $ Ê $ 4 x $ 4. 2 2 Then $ % œ 10 3 or $ œ 3 , or $ 4 œ 5 or $ œ 1; thus $ œ 3 .
10 #
x
10 3
or
10 3
x 5.
kx# 3k !.1 Ê 0.1 x# 3 0.1 Ê 2.9 x# 3.1 Ê È2.9 x È3.1 ¹x È3¹ $ Ê $ x È3 $ Ê $ È3 x $ È3. Then $ È3 œ È2.9 Ê $ œ È3 È2.9 ¸ 0.0291, or $ È3 œ È3.1 Ê $ œ È3.1 È3 ¸ 0.0286; thus $ œ 0.0286.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 2.3 Precise Definition of a Limit 23. Step 1: Step 2:
81
kx# 4k 0.5 Ê 0.5 x# 4 0.5 Ê 3.5 x# 4.5 Ê È3.5 kxk È4.5 Ê È4.5 x È3.5, for x near 2. kx (2)k $ Ê $ x 2 $ Ê $ # x $ 2. Then $ # œ È4.5 Ê $ œ È4.5 # ¸ 0.1213, or $ # œ È3.5 Ê $ œ # È3.5 ¸ 0.1292; thus $ œ È4.5 2 ¸ 0.12.
24. Step 1: Step 2:
25. Step 1: Step 2:
¸ "x (1)¸ 0.1 Ê 0.1
" x
11 1 0.1 Ê 10
" x
9 10 10 10 10 Ê 10 11 x 9 or 9 x 11 .
kx (1)k $ Ê $ x 1 $ Ê $ " x $ ". " 10 " Then $ " œ 10 9 Ê $ œ 9 , or $ " œ 11 Ê $ œ 11 ; thus $ œ
" 11 .
kax# 5b 11k " Ê kx# 16k 1 Ê " x# 16 1 Ê 15 x# 17 Ê È15 x È17. kx 4k $ Ê $ x 4 $ Ê $ % x $ %. Then $ % œ È15 Ê $ œ % È15 ¸ 0.1270, or $ % œ È17 Ê $ œ È17 % ¸ 0.1231; thus $ œ È17 4 ¸ 0.12.
26. Step 1: Step 2:
27. Step 1: Step 2:
28. Step 1: Step 2:
29. Step 1: Step 2:
¸ 120 ¸ x 5 " Ê "
Step 2:
&1 Ê 4
120 x
6 Ê
" 4
x 120
" 6
Ê 30 x 20 or 20 x 30.
kx 24k $ Ê $ x 24 $ Ê $ 24 x $ 24. Then $ 24 œ 20 Ê $ œ 4, or $ 24 œ 30 Ê $ œ 6; thus Ê $ œ 4. kmx 2mk 0.03 Ê 0.03 mx 2m 0.03 Ê 0.03 2m mx 0.03 2m Ê 0.03 2 0.03 m x2 m . kx 2k $ Ê $ x 2 $ Ê $ # x $ #. 0.03 0.03 Then $ # œ # 0.03 m Ê $ œ m , or $ # œ # m Ê $ œ
0.03 m .
In either case, $ œ
kmx 3mk c Ê c mx 3m c Ê c 3m mx c 3m Ê 3 kx 3k $ Ê $ x 3 $ Ê $ $ B $ $. Then $ $ œ $ mc Ê $ œ mc , or $ $ œ $ mc Ê $ œ ¸(mx b) ˆ m# b‰¸ - Ê c mx m# c Ê c ¸x "# ¸ $ Ê $ x "# $ Ê $ "# x $ "# . Then $
30. Step 1:
120 x
" #
œ
" #
c m
Ê $œ
c m,
or $
" #
œ
" #
c m
c m. m #
Ê $œ
c m
x 3
In either case, $ œ
c m.
In either case, $ œ
c m.
m #
Ê
c m
c m. " #
mx c
0.03 m .
c m
x
" #
c m.
k(mx b) (m b)k 0.05 Ê 0.05 mx m 0.05 Ê 0.05 m mx 0.05 m 0.05 Ê 1 0.05 m x" m . kx 1k $ Ê $ x 1 $ Ê $ " x $ ". 0.05 0.05 Then $ " œ " 0.05 m Ê $ œ m , or $ " œ " m Ê $ œ
0.05 m .
In either case, $ œ
0.05 m .
31. lim (3 2x) œ 3 2(3) œ 3 xÄ3
Step 1: Step 2:
32.
ka3 2xb (3)k 0.02 Ê 0.02 6 2x 0.02 Ê 6.02 2x 5.98 Ê 3.01 x 2.99 or 2.99 x 3.01. 0 k x 3k $ Ê $ x 3 $ Ê $ $ x $ $ . Then $ $ œ 2.99 Ê $ œ 0.01, or $ $ œ 3.01 Ê $ œ 0.01; thus $ œ 0.01.
lim (3x #) œ (3)(1) 2 œ 1
x Ä 1
Step 1:
k(3x 2) 1k 0.03 Ê 0.03 3x 3 0.03 Ê 0.01 x 1 0.01 Ê 1.01 x 0.99.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
82
Chapter 2 Limits and Continuity kx (1)k $ Ê $ x 1 $ Ê $ " x $ 1. Then $ " œ 1.01 Ê $ œ 0.01, or $ " œ 0.99 Ê $ œ 0.01; thus $ œ 0.01.
Step 2:
33. lim
x# 4
x Ä # x#
34.
35.
œ lim
xÄ#
#
(x 2)(x 2) (x 2)
œ lim (x 2) œ # # œ 4, x Á 2 xÄ#
(x 2)(x 2) (x 2)
Step 1:
¹Š xx 24 ‹
Step 2:
Ê 1.95 x 2.05, x Á 2. kx 2k $ Ê $ x 2 $ Ê $ # x $ 2. Then $ # œ 1.95 Ê $ œ 0.05, or $ # œ 2.05 Ê $ œ 0.05; thus $ œ 0.05.
lim
x Ä &
x# 6x 5 x5
4¹ 0.05 Ê 0.05
œ lim
x Ä &
(x 5)(x 1) (x 5)
% 0.05 Ê 3.95 x 2 4.05, x Á 2
œ lim (x 1) œ 4, x Á 5. x Ä &
(x 5)(x ") (x 5)
Step 1:
# ¹Š x x 6x5 5 ‹
Step 2:
Ê 5.05 x 4.95, x Á 5. kx (5)k $ Ê $ x 5 $ Ê $ & x $ &. Then $ & œ 5.05 Ê $ œ 0.05, or $ & œ 4.95 Ê $ œ 0.05; thus $ œ 0.05.
(4)¹ 0.05 Ê 0.05
4 0.05 Ê 4.05 x 1 3.95, x Á 5
lim È1 5x œ È1 5(3) œ È16 œ 4
x Ä $
Step 1:
¹È1 5x 4¹ 0.5 Ê 0.5 È1 5x 4 0.5 Ê 3.5 È1 5x 4.5 Ê 12.25 1 5x 20.25
Step 2:
Ê 11.25 5x 19.25 Ê 3.85 x 2.25. kx (3)k $ Ê $ x 3 $ Ê $ $ x $ $. Then $ $ œ 3.85 Ê $ œ 0.85, or $ $ œ 2.25 Ê 0.75; thus $ œ 0.75.
36. lim
4
xÄ# x
œ
4 #
œ2
Step 1:
¸ 4x 2¸ 0.4 Ê 0.4
Step 2:
kx 2k $ Ê $ x 2 $ Ê $ # x $ #. Then $ # œ 53 Ê $ œ "3 , or $ # œ 5# Ê $ œ "# ; thus $ œ 3" .
4 x
2 0.4 Ê 1.6
4 x
2.4 Ê
10 16
x 4
10 24
Ê
10 4
x
10 6
or
5 3
x 25 .
37. Step 1: Step 2:
k(9 x) 5k % Ê % 4 x % Ê % 4 x % 4 Ê % % x 4 % Ê % % x 4 %. kx 4k $ Ê $ x 4 $ Ê $ % x $ %. Then $ 4 œ % 4 Ê $ œ %, or $ % œ % % Ê $ œ %. Thus choose $ œ %.
38. Step 1:
k(3x 7) 2k % Ê % 3x 9 % Ê 9 % 3x * % Ê 3
Step 2:
39. Step 1: Step 2:
40. Step 1:
% 3
x 3 3% .
kx 3k $ Ê $ x 3 $ Ê $ 3 x $ 3. Then $ 3 œ $ 3% Ê $ œ 3% , or $ 3 œ 3 3% Ê $ œ 3% . Thus choose $ œ 3% . ¹Èx 5 2¹ % Ê % Èx 5 # % Ê # % Èx 5 # % Ê (# %)# x 5 (# %)# Ê (# %)# & x (# %)# 5. kx 9k $ Ê $ x 9 $ Ê $ 9 x $ 9. Then $ * œ %# %% * Ê $ œ %% %# , or $ * œ %# %% * Ê $ œ %% %# . Thus choose the smaller distance, $ œ %% %# . ¹È4 x 2¹ % Ê % È4 x # % Ê # % È4 x # % Ê (# %)# % x (# %)# Ê (# %)# x 4 (# %)# Ê (# %)# % x (# %)# %.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 2.3 Precise Definition of a Limit Step 2:
41. Step 1: Step 2:
42. Step 1: Step 2:
43. Step 1: Step 2:
kx 0k $ Ê $ x $ . Then $ œ (# %)# 4 œ %# %% Ê $ œ %% %# , or $ œ (# %)# 4 œ 4% %# . Thus choose the smaller distance, $ œ 4% %# . For x Á 1, kx# 1k % Ê % x# " % Ê " % x# " % Ê È1 % kxk È1 % Ê È" % x È1 % near B œ ". kx 1k $ Ê $ x 1 $ Ê $ " x $ ". Then $ " œ È1 % Ê $ œ " È1 %, or $ 1 œ È" % Ê $ œ È" % 1. Choose $ œ min š" È1 %ß È1 % "›, that is, the smaller of the two distances. For x Á 2, kx# 4k % Ê % x# 4 % Ê 4 % x# 4 % Ê È4 % kxk È4 % Ê È4 % x È4 % near B œ 2. kx (2)k $ Ê $ x 2 $ Ê $ 2 x $ 2. Then $ 2 œ È% % Ê $ œ È% % #, or $ # œ È% % Ê $ œ # È% %. Choose $ œ min šÈ% % #ß # È% %› . ¸ "x 1¸ % Ê %
" x
"% Ê "%
" x
"% Ê
" 1%
% "%,
" 1%.
x
kx 1k $ Ê $ x 1 $ Ê " $ x " $ . Then " $ œ " " % Ê $ œ " " " % œ " % % , or " $ œ " " % Ê $ œ Choose $ œ
44. Step 1:
83
" "%
"œ
% "%.
the smaller of the two distances.
¸ x"# "3 ¸ % Ê %
" x#
" 3
% Ê
" 3
%
" x#
" 3
% Ê
1 3% 3
" x#
1 $% 3
Ê
3 "
%$ x#
3 "
%$3 È $. Ê É 1 3 $% kxk É " 3 $% , or É " 3 $% x É "$ % for x near
Step 2:
¹x È3¹ $ Ê $ x È3 $ Ê È3 $ x È3 $ . Then È3 $ œ É " 3 $% Ê $ œ È3 É " 3 $% , or È3 $ œ É " 3 $% Ê $ œ É " 3 $% È3. Choose $ œ min šÈ3 É " 3 $% ß É " 3 $% È3›.
45. Step 1: Step 2:
46. Step 1: Step 2:
47. Step 1:
#
¹Š xx*3 ‹ (6)¹ % Ê % (x 3) 6 %, x Á 3 Ê % x 3 % Ê % $ x % $. kx (3)k $ Ê $ x 3 $ Ê $ $ x $ 3. Then $ $ œ % $ Ê $ œ %, or $ $ œ % $ Ê $ œ %. Choose $ œ %. #
¹Š xx11 ‹ 2¹ % Ê % (x 1) 2 %, x Á 1 Ê " % x " %. kx 1k $ Ê $ x 1 $ Ê " $ x " $ . Then " $ œ " % Ê $ œ %, or " $ œ " % Ê $ œ %. Choose $ œ %. x 1: l(4 2x) 2l % Ê ! 2 2x % since x 1Þ Thus, 1
% #
x !;
x 1: l(6x 4) 2l % Ê ! Ÿ 6x 6 % since x 1. Thus, " Ÿ x 1 6% . Step 2:
48. Step 1:
kx 1k $ Ê $ x 1 $ Ê " $ x 1 $ . Then 1 $ œ " #% Ê $ œ #% , or " $ œ 1 6% Ê $ œ 6% . Choose $ œ 6% . x !: k2x 0k % Ê % 2x ! Ê #% x 0; x 0: ¸ x# !¸ % Ê ! Ÿ x #%.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
84
Chapter 2 Limits and Continuity Step 2:
kx 0k $ Ê $ x $ . Then $ œ #% Ê $ œ #% , or $ œ #% Ê $ œ #%. Choose $ œ #% .
49. By the figure, x Ÿ x sin
" x
Ÿ x for all x 0 and x x sin
then by the sandwich theorem, in either case, lim x sin xÄ!
50. By the figure, x# Ÿ x# sin
" x
" x
" x
x for x 0. Since lim (x) œ lim x œ 0, xÄ!
œ 0.
xÄ!
Ÿ x# for all x except possibly at x œ 0. Since lim ax# b œ lim x# œ 0, then
by the sandwich theorem, lim x# sin xÄ!
" x
xÄ!
œ 0.
xÄ!
51. As x approaches the value 0, the values of g(x) approach k. Thus for every number % 0, there exists a $ ! such that ! kx 0k $ Ê kg(x) kk %. 52. Write x œ h c. Then ! lx cl $ Í $ x c $ , x Á c Í $ ah cb c $ , h c Á c Í $ h $ , h Á ! Í ! lh !l $ . Thus, limfaxb œ L Í for any % !, there exists $ ! such that lfaxb Ll % whenever ! lx cl $ xÄc
Í lfah cb Ll % whenever ! lh !l $ Í limfah cb œ L. hÄ!
53. Let f(x) œ x# . The function values do get closer to 1 as x approaches 0, but lim f(x) œ 0, not 1. The xÄ!
function f(x) œ x# never gets arbitrarily close to 1 for x near 0.
54. Let f(x) œ sin x, L œ "# , and x! œ 0. There exists a value of x (namely, x œ 16 ) for which ¸sin x "# ¸ % for any given % 0. However, lim sin x œ 0, not "# . The wrong statement does not require x to be arbitrarily close to xÄ!
x! . As another example, let g(x) œ sin "x , L œ #" , and x! œ 0. We can choose infinitely many values of x near 0 such that sin
" x
œ
" #
as you can see from the accompanying figure. However, lim sin xÄ!
" x
fails to exist. The
wrong statement does not require all values of x arbitrarily close to x! œ 0 to lie within % 0 of L œ "# . Again you can see from the figure that there are also infinitely many values of x near 0 such that sin "x œ 0. If we choose % 4" we cannot satisfy the inequality ¸sin x" #" ¸ % for all values of x sufficiently near x! œ 0.
#
55. kA *k Ÿ 0.01 Ê 0.01 Ÿ 1 ˆ x# ‰ 9 Ÿ 0.01 Ê 8.99 Ÿ Ê
2É 8.99 1
ŸxŸ
2É 9.01 1
1 x# 4
Ÿ 9.01 Ê
4 1
(8.99) Ÿ x# Ÿ
4 1
(9.01)
or 3.384 Ÿ x Ÿ 3.387. To be safe, the left endpoint was rounded up and the right
endpoint was rounded down. 56. V œ RI Ê
V R
œ I Ê ¸ VR 5¸ Ÿ 0.1 Ê 0.1 Ÿ
120 R
5 Ÿ 0.1 Ê 4.9 Ÿ
120 R
Ÿ 5.1 Ê
10 49
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
R 1#0
10 51
Ê
Section 2.3 Precise Definition of a Limit (120)(10) 51
ŸRŸ
(120)(10) 49
85
Ê 23.53 Ÿ R Ÿ 24.48.
To be safe, the left endpoint was rounded up and the right endpoint was rounded down. 57. (a) $ x 1 0 Ê " $ x 1 Ê f(x) œ x. Then kf(x) 2k œ kx 2k œ 2 x 2 1 œ 1. That is, kf(x) 2k 1 "# no matter how small $ is taken when " $ x 1 Ê lim f(x) Á 2. xÄ1
(b) 0 x 1 $ Ê " x " $ Ê f(x) œ x 1. Then kf(x) 1k œ k(x 1) 1k œ kxk œ x 1. That is, kf(x) 1k 1 no matter how small $ is taken when " x " $ Ê lim f(x) Á 1. xÄ1
(c) $ x 1 ! Ê " $ x 1 Ê f(x) œ x. Then kf(x) 1.5k œ kx 1.5k œ 1.5 x 1.5 1 œ 0.5. Also, ! x 1 $ Ê 1 x " $ Ê f(x) œ x 1. Then kf(x) 1.5k œ k(x 1) 1.5k œ kx 0.5k œ x 0.5 " 0.5 œ 0.5. Thus, no matter how small $ is taken, there exists a value of x such that $ x 1 $ but kf(x) 1.5k "# Ê lim f(x) Á 1.5. xÄ1
58. (a) For 2 x 2 $ Ê h(x) œ 2 Ê kh(x) 4k œ 2. Thus for % 2, kh(x) 4k % whenever 2 x 2 $ no matter how small we choose $ 0 Ê lim h(x) Á 4. xÄ#
(b) For 2 x 2 $ Ê h(x) œ 2 Ê kh(x) 3k œ 1. Thus for % 1, kh(x) 3k % whenever 2 x 2 $ no matter how small we choose $ 0 Ê lim h(x) Á 3. xÄ#
(c) For 2 $ x 2 Ê h(x) œ x# so kh(x) 2k œ kx# 2k . No matter how small $ 0 is chosen, x# is close to 4 when x is near 2 and to the left on the real line Ê kx# 2k will be close to 2. Thus if % 1, kh(x) 2k % whenever 2 $ x 2 no mater how small we choose $ 0 Ê lim h(x) Á 2. xÄ#
59. (a) For 3 $ x 3 Ê f(x) 4.8 Ê kf(x) 4k 0.8. Thus for % 0.8, kf(x) 4k % whenever 3 $ x 3 no matter how small we choose $ 0 Ê lim f(x) Á 4. xÄ$
(b) For 3 x 3 $ Ê f(x) 3 Ê kf(x) 4.8k 1.8. Thus for % 1.8, kf(x) 4.8k % whenever 3 x 3 $ no matter how small we choose $ 0 Ê lim f(x) Á 4.8. xÄ$
(c) For 3 $ x 3 Ê f(x) 4.8 Ê kf(x) 3k 1.8. Again, for % 1.8, kf(x) 3k % whenever $ $ x 3 no matter how small we choose $ 0 Ê lim f(x) Á 3. xÄ$
60. (a) No matter how small we choose $ 0, for x near 1 satisfying " $ x " $ , the values of g(x) are near 1 Ê kg(x) 2k is near 1. Then, for % œ "# we have kg(x) 2k "# for some x satisfying " $ x " $ , or ! kx 1k $ Ê
lim g(x) Á 2.
x Ä 1
(b) Yes, lim g(x) œ 1 because from the graph we can find a $ ! such that kg(x) 1k % if ! kx (1)k $ . x Ä 1
61-66. Example CAS commands (values of del may vary for a specified eps): Maple: f := x -> (x^4-81)/(x-3);x0 := 3; plot( f(x), x=x0-1..x0+1, color=black, # (a) title="Section 2.3, #61(a)" ); L := limit( f(x), x=x0 ); # (b) epsilon := 0.2; # (c) plot( [f(x),L-epsilon,L+epsilon], x=x0-0.01..x0+0.01, color=black, linestyle=[1,3,3], title="Section 2.3, #61(c)" ); q := fsolve( abs( f(x)-L ) = epsilon, x=x0-1..x0+1 ); # (d) delta := abs(x0-q); plot( [f(x),L-epsilon,L+epsilon], x=x0-delta..x0+delta, color=black, title="Section 2.3, #61(d)" ); for eps in [0.1, 0.005, 0.001 ] do # (e)
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86
Chapter 2 Limits and Continuity q := fsolve( abs( f(x)-L ) = eps, x=x0-1..x0+1 ); delta := abs(x0-q); head := sprintf("Section 2.3, #61(e)\n epsilon = %5f, delta = %5f\n", eps, delta ); print(plot( [f(x),L-eps,L+eps], x=x0-delta..x0+delta, color=black, linestyle=[1,3,3], title=head )); end do: Mathematica (assigned function and values for x0, eps and del may vary): Clear[f, x] y1: œ L eps; y2: œ L eps; x0 œ 1; f[x_]: œ (3x2 (7x 1)Sqrt[x] 5)/(x 1) Plot[f[x], {x, x0 0.2, x0 0.2}] L: œ Limit[f[x], x Ä x0] eps œ 0.1; del œ 0.2; Plot[{f[x], y1, y2},{x, x0 del, x0 del}, PlotRange Ä {L 2eps, L 2eps}]
2.4 ONE-SIDED LIMITS AND LIMITS AT INFINITY 1. (a) True (e) True (i) False
(b) True (f) True (j) False
(c) False (g) False (k) True
(d) True (h) False (l) False
2. (a) True (e) True (i) True
(b) False (f) True (j) False
(c) False (g) True (k) True
(d) True (h) True
3. (a)
lim f(x) œ
x Ä #b
2 #
" œ #, lim c f(x) œ $ # œ " xÄ#
(b) No, lim f(x) does not exist because lim b f(x) Á lim c f(x) xÄ# xÄ# xÄ# (c) lim c f(x) œ 4# 1 œ 3, lim b f(x) œ 4# " œ $ xÄ%
xÄ%
(d) Yes, lim f(x) œ 3 because 3 œ lim c f(x) œ lim b f(x) xÄ% xÄ% xÄ% 4. (a)
lim f(x) œ
x Ä #b
2 #
œ 1, lim c f(x) œ $ # œ ", f(2) œ 2 xÄ#
(b) Yes, lim f(x) œ 1 because " œ lim b f(x) œ lim c f(x) xÄ# xÄ# xÄ# (c) lim c f(x) œ 3 (1) œ 4, lim b f(x) œ 3 (1) œ 4 x Ä "
x Ä "
(d) Yes, lim f(x) œ 4 because 4 œ x Ä "
lim
x Ä "c
f(x) œ
lim
x Ä "b
f(x)
5. (a) No, lim b f(x) does not exist since sin ˆ "x ‰ does not approach any single value as x approaches 0 xÄ! (b) lim c f(x) œ lim c 0 œ 0 xÄ!
(c)
xÄ!
lim f(x) does not exist because lim b f(x) does not exist xÄ! xÄ!
6. (a) Yes, lim b g(x) œ 0 by the sandwich theorem since Èx Ÿ g(x) Ÿ Èx when x 0 xÄ! (b) No, lim c g(x) does not exist since Èx is not defined for x 0 xÄ!
(c) No, lim g(x) does not exist since lim c g(x) does not exist xÄ! xÄ!
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 2.4 One-Sided Limits and Limits at Infinity 7. (aÑ
lim f(x) œ " œ lim b f(x) xÄ1 (c) Yes, lim f(x) œ 1 since the right-hand and left-hand (b)
x Ä 1c
xÄ1
limits exist and equal 1
8. (a)
(b)
lim f(x) œ 0 œ lim c f(x) xÄ1
x Ä 1b
(c) Yes, lim f(x) œ 0 since the right-hand and left-hand xÄ1
limits exist and equal 0
9. (a) domain: 0 Ÿ x Ÿ 2 range: 0 y Ÿ 1 and y œ 2 (b) xlim f(x) exists for c belonging to Äc (0ß 1) ("ß #) (c) x œ 2 (d) x œ 0
10. (a) domain: _ x _ range: " Ÿ y Ÿ 1 (b) xlim f(x) exists for c belonging to Äc (_ß 1) ("ß ") ("ß _) (c) none (d) none
11.
x Ä !Þ&c
lim
13.
x Ä #b
14.
x Ä 1c
15.
lim b
lim
lim
hÄ!
2 0.5 2 È3 É 3/2 É xx É 1 œ 0.5 1 œ 1/2 œ
12.
lim
x Ä 1b
" 1 È0 œ ! É "1 É xx # œ # œ
5‰ ˆ x x 1 ‰ ˆ 2x ˆ 2 ‰ 2(2) 5 ˆ"‰ x# x œ # " Š (#)# (2) ‹ œ (2) # œ 1
ˆ x " 1 ‰ ˆ x x 6 ‰ ˆ 3 7 x ‰ œ ˆ 1 " 1 ‰ ˆ 1 1 6 ‰ ˆ 3 7 1 ‰ œ ˆ "# ‰ ˆ 71 ‰ ˆ 27 ‰ œ 1 Èh# 4h 5 È5 h
œ lim b hÄ!
œ lim b Š hÄ!
ah# 4h 5b 5 h ŠÈh# 4h 5 È5‹
Èh# 4h 5 È5 È # 4h 5 È5 ‹ Š Èhh# ‹ h 4h 5 È5
œ lim b hÄ!
h(h 4) h ŠÈh# 4h 5 È5‹
œ
04 È5 È5
œ
2 È5
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
87
88 16.
Chapter 2 Limits and Continuity lim
h Ä !c
È6 È5h# 11h 6 h
6 a5h# 11h 6b
œ lim c hÄ! 17. (a)
19. (a)
) Ä $b
20. (a)
t Ä %b
(x2) (x#)
Ú) Û )
lim
akx 2k œ x 2 for x 2b
(x 3) ’ (x(x#2) ) “
lim
x Ä #c
akx 2k œ (x 2) for x 2b
(x 3)(1) œ (2 3) œ 1
È2x (x 1) (x 1)
akx 1k œ x 1 for x 1b
œ lim b È2x œ È2 xÄ1
œ lim c xÄ1
È2x (x 1) (x 1)
akx 1k œ (x 1) for x 1b
œ
œ1
3 3
lim at ÚtÛb œ 4 4 œ 0 sin È2) È 2)
22. lim
sin kt t
23. lim
sin 3y 4y
)Ä!
tÄ!
yÄ!
œ
26. lim
2t
t Ä ! tan t
27. lim
xÄ!
)Ä!
3 sin 3y " 4 ylim 3y Ä!
sin 2x ‰ ˆ cos 2x x
œ lim
xÄ!
œ 2 lim
t
sin t t Ä ! ˆ cos t ‰
x csc 2x cos 5x
œ
œ
œ lim
tÄ!
" ‰ cos 5x
29. lim
x x cos x
) Ä $c
(b)
t Ä %c
Ú) Û )
lim
œ
2 3
lim at ÚtÛb œ 4 3 œ 1
t cos t sin t
œ lim ˆ sin xxcos x xÄ!
œk†1œk
3 sin ) 4 )lim Ä! )
œ
" 3
Œ
œ
" lim
)Ä!c
(where ) œ kt) (where ) œ 3y)
3 4
œ
sin ) )
"
œ Š lim
‹ Š lim x Ä ! cos 2x xÄ!
sin 2x
" 3
†1œ
2 sin 2x #x ‹
" 3
(where ) œ 3h)
œ1†2œ2
œ 2 Š lim cos t‹ Œ lim" sin t œ 2 † " † " œ 2 tÄ!
œ Š #" lim
t
Ä!
t
"
‹ Š lim cos 5x ‹ x Ä ! sin 2x xÄ!
6x# cos x sin x sin 2x xÄ!
x Ä ! sin x cos x
œ
" " sin 3h 3 h lim Ä !c ˆ 3h ‰
28. lim 6x# (cot x)(csc 2x) œ lim xÄ!
)Ä!
x Ä ! x cos 2x
œ 2 lim
sin ) )
œ k lim
sin 3y 3 4 ylim Ä ! 3y
3h ‰ sin 3h
œ lim ˆ sinx2x † xÄ!
k sin ) )
œ lim
œ lim c ˆ "3 † hÄ!
h
tan 2x x
k sin kt kt
(b)
(where x œ È2))
œ1
sin x x
xÄ!
tÄ!
lim h Ä !c sin 3h
xÄ!
œ lim
œ lim
25. lim
œ 211 È6
œ lim c È2x œ È2 xÄ1
21. lim
24.
lim
(0 11) È6 È6
œ
(x 3) œ (2) 3 œ 1
lim
x Ä #b
x Ä #c
œ lim b xÄ1
È2x (x 1) kx 1 k
lim
x Ä 1c
œ
h(5h 11)
h ŠÈ6 È5h# 11h 6‹
(x 3)
lim
x Ä #b
œ
È2x (x 1) kx 1 k
lim
(b)
kx 2 k x2
(x 3)
lim
x Ä 1b
œ
œ
x Ä #c
18. (a)
kx 2 k x 2
(x 3)
lim
È5h# 11h 6 È6 È5h# 11h 6 È ‹ Š È66 ‹ h È5h# 11h 6
œ lim c hÄ!
h ŠÈ6 È5h# 11h 6‹
x Ä #b
(b)
œ lim c Š hÄ!
2x
œ lim ˆ3 cos x † xÄ!
x cos x ‰ sin x cos x
œ lim ˆ sinx x † xÄ!
x sin x
†
" ‰ cos x
œ ˆ #" † 1‰ (1) œ
2x ‰ sin 2x
" #
œ3†"†1œ3
lim
x
x Ä ! sin x
œ lim Š sin" x ‹ † lim ˆ cos" x ‰ lim Š sin" x ‹ œ (1)(1) 1 œ 2 xÄ!
30. lim
xÄ!
x
x# x sin x #x
xÄ!
œ lim ˆ #x xÄ!
xÄ!
" #
x
"# ˆ sinx x ‰‰ œ 0
" #
"# (1) œ 0
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 2.4 One-Sided Limits and Limits at Infinity 31. lim
sin(1 cos t) 1cos t
32. lim
sin (sin h) sin h
tÄ!
hÄ!
sin )
33. lim
) Ä ! sin 2)
34. lim
sin 5x
35. lim
tan 3x
œ
3 8 xlim Ä!
36. lim
yÄ!
œ
)Ä!
sin ) )
sin ) )
œ lim
)Ä!
œ 1 since ) œ 1 cos t Ä 0 as t Ä 0
œ 1 since ) œ sin h Ä 0 as h Ä 0
sin ) œ lim ˆ sin 2) †
2) ‰ #)
5x œ lim ˆ sin sin 4x †
4x 5x
sin 3x œ lim ˆ cos 3x †
" ‰ sin 8x
)Ä!
x Ä ! sin 4x
x Ä ! sin 8x
œ lim
xÄ!
xÄ!
" # )lim Ä!
œ
† 54 ‰ œ
œ lim
yÄ!
2) ‰ sin 2)
ˆ sin5x5x †
5 4 xlim Ä!
sin 3x œ lim ˆ cos 3x † 3 8
†1†1†1œ
4x ‰ sin 4x
†
8x 3x
†1†1œ œ
5 4
3 8
"
lim
xÄ „_
12 5
œ
12 5
œ 0 whenever
m n
0. This result follows immediately from
ˆ xm"În ‰ œ
lim
xÄ „_
37. (a) 3
(b) 3
38. (a) 1
(b) 1
39. (a)
" #
(b)
" #
40. (a)
" 8
(b)
" 8
41. (a) 53
45.
lim
tÄ_
46. r Ä lim_
ˆ x" ‰mÎn œ Š
"
lim ‹ xÄ „_ x
mÎn
(b) 53
3 4
(b)
44. 3") Ÿ
5 4
† 83 ‰
œ1†1†1†1†
lim mÎn xÄ „_ x
Example 6 and the power rule in Theorem 8:
43. "x Ÿ
†1†1œ
yÄ!
cos 5y ˆ 3†4 ‰ lim Š sin3y3y ‹ Š sin4y4y ‹ Š sin5y5y ‹ Š cos 4y ‹ 5 yÄ!
42. (a)
" #
sin 4y cos 5y 3†4†5y œ lim Š siny3y ‹ Š cos 4y ‹ Š sin 5y ‹ Š 3†4†5y ‹
sin 3y sin 4y cos 5y y cos 4y sin 5y
Note: In these exercises we use the result
" #
œ
" sin 8x
xÄ!
ˆ cos"3x ‰ ˆ sin3x3x ‰ ˆ sin8x8x ‰ œ
sin 3y cot 5y y cot 4y
ˆ sin) ) †
sin 2x x
Ÿ
" x
cos ) 3)
Ÿ
" 3)
2 t sin t t cos t
Ê x lim Ä_ Ê
47. (a) x lim Ä_
lim
) Ä _
œ lim
2 t
tÄ_
r sin r 2r 7 5 sin r
2x 3 5x 7
$
sin 2x x
œrÄ lim_
œ x lim Ä_
œ 0 by the Sandwich Theorem
cos ) 3)
œ 0 by the Sandwich Theorem
1 ˆ sint t ‰ 1 ˆ cost t ‰
œ
1 ˆ sinr r ‰ 2 7r 5 ˆ sinr r ‰ 2 3x 5 7x
2x 7 48. (a) x lim œ x lim Ä _ x$ x# x 7 Ä_ (b) 2 (same process as part (a))
3 4
œ
010 10
œ 1
œrÄ lim_
2 5
2 Š x7$ ‹
1 "x x"# x7$
10 200
œ
(b)
" #
2 5
(same process as part (a))
œ2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
œ 0mÎn œ 0.
89
90
Chapter 2 Limits and Continuity " x
x"#
49. (a) x lim Ä_
x1 x# 3
œ x lim Ä_
1 x3#
50. (a) x lim Ä_
3x 7 x# 2
œ x lim Ä_
1 x2#
51. (a) x lim Ä_
7x$ x$ 3x# 6x
52. (a) x lim Ä_
" x$ 4x 1
&
3 x
x7#
œ x lim Ä_
(b)
9 #
2x%
œ0
(b) 0 (same process as part (a))
" x$
œ x lim Ä_
10 x
œ x lim Ä_
œ(
x"# x31' 1
2
(b) 7 (same process as part (a))
œ!
1 x4# x"$
%
9x% x 5x# x 6
(b) 0 (same process as part (a))
7 1 3x x6#
10x x 31 53. (a) x lim œ x lim x' Ä_ Ä_ (b) 0 (same process as part (a))
54. (a) x lim Ä_
œ0
(b) 0 (same process as part (a))
œ0
9 x"$
5 x#
x"$ x6%
œ
9 #
(same process as part (a))
55. (a) x lim Ä_
2x$ 2x 3 3x$ 3x# 5x
œ x lim Ä_
2 x2# x3$ 3 3x x5#
œ 23
(b) 23 (same process as part (a)) %
x 56. (a) x lim œ x lim Ä _ x% 7x$ 7x# 9 Ä_ (b) 1 (same process as part (a))
57. x lim Ä_
2Èx x" 3x 7
59. x Ä lim _
œ x lim Ä_
$ & x È xÈ $ xÈ & x È
x" x% x# x$
61. x lim Ä_
2x&Î$ x"Î$ 7 x)Î& 3x Èx
62. x Ä lim _
2 Š "Î# ‹ Š x"# ‹ x 3 7x
œxÄ lim _
60. x lim Ä_
œ x lim Ä_
$ È x 5x 3 2x x#Î$ 4
" 1 7x x7# x9%
œ0
1 xÐ"Î&Ñ Ð"Î$Ñ 1 xÐ"Î&Ñ Ð"Î$Ñ
x x"# 1 x"
œ x lim Ä_ œxÄ lim _
œ 1
58. x lim Ä_
œxÄ lim _
" ‹ 1 Š #Î"& x
" ‹ 1 Š #Î"&
2 Èx 2 Èx
œ x lim Ä_
2 Š "Î# ‹" x 2 Š "Î# ‹1 x
œ 1
œ1
x
œ_
" 7 2x"Î"& "*Î"& x x)Î& " 3 1 $Î& ""Î"! x
" x#Î$
2
5 3x " x"Î$
œ_
x
4x
œ 5#
63. Yes. If lim b f(x) œ L œ lim c f(x), then xlim f(x) œ L. If lim b f(x) Á lim c f(x), then xlim f(x) does not exist. Äa Äa xÄa xÄa xÄa xÄa 64. Since xlim f(x) œ L if and only if lim b f(x) œ L and lim c f(x) œ L, then xlim f(x) can be found by calculating Äc Äc xÄc xÄc lim b f(x). xÄc
65. If f is an odd function of x, then f(x) œ f(x). Given lim b f(x) œ 3, then lim c f(x) œ $. xÄ! xÄ!
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 2.4 One-Sided Limits and Limits at Infinity
91
66. If f is an even function of x, then f(x) œ f(x). Given lim c f(x) œ 7 then lim b f(x) œ 7. However, nothing xÄ# x Ä # can be said about 67. Yes. If x lim Ä_
lim
x Ä #c
f(x) g(x)
f(x) because we don't know lim b f(x). xÄ#
œ 2 then the ratio of the polynomials' leading coefficients is 2, so x Ä lim _
f(x) g(x)
œ 2 as well.
68. Yes, it can have a horizontal or oblique asymptote. 69. At most 1 horizontal asymptote: If x lim Ä_ f(x) lim x Ä _ g(x)
f(x) g(x)
œ L, then the ratio of the polynomials' leading coefficients is L, so
œ L as well.
Èx# x Èx# x œ lim ’Èx# x Èx# x“ † ’ Èx# x Èx# x “ œ lim 70. x lim È x# x È x# x Ä_ xÄ_ xÄ_ 2x 2 2 œ x lim œ lim œ œ 1 È # 1 1 " " Ä_ È # xÄ_ x x
x x
ax # x b a x # x b È x# x È x# x
É1 x É1 x
71. For any % 0, take N œ 1. Then for all x N we have that kf(x) kk œ kk kk œ 0 %. 72. For any % 0, take N œ 1. Then for all y N we have that kf(x) kk œ kk kk œ 0 %. 73. I œ (5ß 5 $ ) Ê 5 x & $ . Also, Èx 5 % Ê x 5 %# Ê x & %# . Choose $ œ %# Ê lim Èx 5 œ 0. x Ä &b
74. I œ (% $ ß %) Ê % $ x 4. Also, È% x % Ê % x %# Ê x % %# . Choose $ œ %# Ê lim È% x œ 0. x Ä %c
75. As x Ä 0 the number x is always negative. Thus, ¹ kxxk (1)¹ % Ê ¸ xx 1¸ % Ê 0 % which is always true independent of the value of x. Hence we can choose any $ 0 with $ x ! Ê
x
lim x Ä ! c kx k
œ 1.
2 ¸ x 2 ¸ 76. Since x Ä # we have x 2 and kx 2k œ x 2. Then, ¹ kxx 2 k " ¹ œ x 2 " % Ê 0 %
which is always true so long as x #. Hence we can choose any $ !, and thus # x # $ 2 Ê ¹ kxx 2k "¹ % . Thus,
77. (a)
lim
x Ä %!!b
x 2
lim x Ä #b kx2k
œ 1.
ÚxÛ œ 400. Just observe that if 400 x 401, then ÚxÛ œ 400. Thus if we choose $ œ ", we have for any
number % ! that 400 x 400 $ Ê lÚxÛ 400l œ l400 400l œ ! %. (b) lim c ÚxÛ œ 399. Just observe that if 399 x 400 then ÚxÛ œ 399. Thus if we choose $ œ ", we have for any x Ä %!!
number % ! that 400 $ x 400 Ê lÚxÛ 399l œ l399 399l œ ! %. (c) Since lim b ÚxÛ Á lim c ÚxÛ we conclude that lim ÚxÛ does not exist. x Ä %!!
78. (a)
x Ä %!!
x Ä %!!
lim f(x) œ lim b Èx œ È0 œ 0; ¸Èx 0¸ % Ê % Èx % Ê ! x %# for x positive. Choose $ œ %# xÄ! Ê lim b f(x) œ 0.
x Ä !b
xÄ!
(b)
lim f(x) œ lim c x# sin ˆ x" ‰ œ 0 by the sandwich theorem since x# Ÿ x# sin ˆ x" ‰ Ÿ x# for all x Á 0. x Ä !c xÄ! Since kx# 0k œ kx# 0k œ x# % whenever kxk È%, we choose $ œ È% and obtain ¸x# sin ˆ "x ‰ 0¸ % if $ x 0.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
92
Chapter 2 Limits and Continuity (c) The function f has limit 0 at x! œ 0 since both the right-hand and left-hand limits exist and equal 0.
79.
81.
82. 83. 84.
lim
" x
x sin
xÄ „_
3x 4 x Ä „ _ 2x 5
œ lim
œ
lim
"
)Ä0 )
sin ) œ 1, ˆ) œ x" ‰
3 4x 5 x Ä „ _ 2 x
lim
œ lim
3 4t
t Ä 0 2 5t
œ
80. 3 #
cos
lim
" x
x Ä _ 1 x"
œ lim c )Ä!
cos ) 1)
œ
" 1
œ 1, ˆ) œ x" ‰
, ˆt œ "x ‰
"Îx lim ˆ "x ‰ œ lim b zz œ 1, ˆz œ x" ‰ zÄ!
xÄ_
ˆ3 2x ‰ ˆcos "x ‰ œ lim (3 2))(cos )) œ (3)(1) œ 3, ˆ) œ x" ‰
lim
xÄ „_
)Ä0
lim ˆ x3# cos x" ‰ ˆ1 sin x" ‰ œ lim b a3)# cos )b (1 sin )) œ (0 1)(1 0) œ 1, ˆ) œ x" ‰ )Ä!
xÄ_
2.5 INFINITE LIMITS AND VERTICAL ASYMPTOTES "
œ_
1.
lim x Ä !b 3x
3.
lim x Ä #c x 2
5.
lim x Ä )b x8
7.
lim # x Ä ( (x7)
3
2x
4
œ _ œ _ œ_
lim "Î$ x Ä !b 3x
10. (a)
lim "Î& x Ä !b x 4
11. lim
#Î& xÄ! x
13.
œ lim
4
# x Ä ! ax"Î& b
œ_
Š positive positive ‹
lim x Ä !c 2x
positive Š negative ‹
4.
lim x Ä $b x 3
Š negative positive ‹
6.
lim x Ä &c 2x10
3x
œ_
positive Š positive ‹
8.
lim # x Ä ! x (x1)
œ _
(b)
lim "Î$ x Ä !c 3x
(b)
lim "Î& x Ä !c x
œ_
2
positive Š negative ‹
2.
œ_
2
9. (a)
œ _
Š positive positive ‹
œ_
5
"
"
2
2
12. lim
"
#Î$ xÄ! x
lim tan x œ _
14.
x Ä ˆ 1# ‰
Š negative negative ‹ negative Š positive †positive ‹
œ _ œ _
œ lim
"
# x Ä ! ax"Î$ b
œ_
lim sec x œ _
x Ä ˆ #1 ‰
lim (1 csc )) œ _
15.
) Ä !
16.
) Ä !b
lim (2 cot )) œ _ and lim c (2 cot )) œ _, so the limit does not exist )Ä!
"
œ lim b xÄ#
" (x2)(x2)
œ_
Š positive"†positive ‹
"
œ lim c xÄ#
" (x2)(x2)
œ _
Š positive†"negative ‹
17. (a)
lim # x Ä # b x 4
(b)
lim # x Ä # c x 4
(c)
lim # x Ä #b x 4
(d)
lim # x Ä #c x 4
"
œ
lim x Ä #b (x2)(x2)
"
œ _
Š positive†"negative ‹
"
œ
lim x Ä #c (x2)(x2)
"
œ_
Š negative"†negative ‹
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 2.5 Infinite Limits and Vertical Asymptotes x
œ lim b xÄ"
x (x1)(x1)
œ_
positive Š positive †positive ‹
x
œ lim c xÄ"
x (x1)(x1)
œ _
positive Š positive †negative ‹
18. (a)
lim # x Ä "b x 1
(b)
lim # x Ä "c x 1
(c)
lim # x Ä "b x 1
(d)
lim # x Ä "c x 1
x
œ
lim x Ä "b (x1)(x1)
x
œ_
negative Š positive †negative ‹
x
œ
lim x Ä "c (x1)(x1)
x
œ _
negative Š negative †negative ‹
19. (a)
lim x Ä !b #
x#
" x
œ 0 lim b xÄ!
" x
œ _
" Š negative ‹
(b)
lim x Ä !c #
x#
" x
œ 0 lim c xÄ!
" x
œ_
" Š positive ‹
(c)
lim # x Ä $È2
(d)
lim x Ä 1 #
20. (a)
x#
x#
lim x Ä #b
" x
" x
œ
x# 1 2x 4 x# 1
2#Î$ #
œ
" #
(d)
lim x Ä !c 2x 4
œ
(b) (c) (d) (e) 22. (a)
x# 3x 2 x$ 2x#
lim b
x# 3x 2 x$ 2x# #
x 3x 2 x$ 2x#
lim
x Ä #c
lim
xÄ#
#
x 3x 2 x$ 2x# #
x 3x 2 x$ 2x#
lim
xÄ!
lim x Ä #b
(c)
x Ä 0c
œ lim b xÄ# œ lim c xÄ#
(d)
x Ä "b
(e)
lim x Ä !b x(x #)
x# 1 2x 4
lim x Ä #c
œ _
positive Š negative ‹
œ0
(x 2)(x 1) x# (x 2)
œ _
(x 2)(x 1) x# (x 2)
œ lim b xÄ#
(x 2)(x 1) x# (x 2)
œ lim c xÄ#
œ lim
œ lim
(x 2)(x 1) x# (x 2)
œ _
xÄ!
x# 3x 2 x$ 4x
lim
x# 3x 2 x$ 4x x"
2†0 #4
(x 2)(x 1) x# (x 2)
xÄ#
lim
and
œ lim b xÄ!
œ lim b xÄ#
x# 3x 2 x$ 4x
(b)
œ
œ lim
x# 3x 2 x$ 4x
lim
x Ä #b
(x 1)(x 1) 2x 4
(b)
" 4
lim b
xÄ#
3 #
Š positive positive ‹
œ lim b xÄ"
xÄ!
œ 2"Î$ 2"Î$ œ 0
œ_
lim x Ä "b 2x 4
21. (a)
" #"Î$
ˆ "1 ‰ œ
(c)
x# 1
œ
xÄ#
(x 2)(x ") x(x #)(x 2)
(x 2)(x ")
œ lim c xÄ! œ lim b xÄ"
(x 2)(x ") x(x #)(x 2) (x 2)(x ") x(x #)(x 2)
œ
x1 x#
œ
" 4
,xÁ2
x1 x#
œ
" 4
,xÁ2
x1 x#
œ
" 4
,xÁ2 †negative Š negative positive†negative ‹
œ lim b xÄ#
lim x Ä #b x(x #)(x 2)
†negative Š negative positive†negative ‹
(x 1) x(x #)
œ
(x 1)
lim x Ä #b x(x #)
œ lim c xÄ! œ lim b xÄ"
œ_
(x 1) x(x #)
œ
negative Š positive †positive ‹
x"
negative Š negative †positive ‹
œ_
œ
" 8
œ_
(x 1) x(x #)
œ _
lim x Ä !c x(x #)
" #(4)
0 (1)(3)
negative Š negative †positive ‹ negative Š negative †positive ‹
œ0
so the function has no limit as x Ä 0. lim 2
23. (a)
t Ä !b
24. (a)
t Ä !b
25. (a)
x Ä !b
(c)
x Ä "b
3 ‘ t"Î$
œ _
" lim t$Î& 7‘ œ _
lim
2
lim
lim
" ’ x#Î$
2 “ (x 1)#Î$
œ_
lim
" ’ x#Î$
2 “ (x 1)#Î$
œ_
(b)
t Ä !c
(b)
t Ä !c
lim
" ’ x#Î$
2 “ (x 1)#Î$
œ_
(b)
x Ä !c
lim
" ’ x#Î$
2 “ (x 1)#Î$
œ_
(d)
x Ä "c
" t$Î&
3 ‘ t"Î$
œ_
7‘ œ _
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93
94
Chapter 2 Limits and Continuity
26. (a)
x Ä !b
(c)
x Ä "b
lim
" ’ x"Î$
1 “ (x 1)%Î$
œ_
(b)
x Ä !c
lim
" ’ x"Î$
1 “ (x 1)%Î$
œ _
lim
" ’ x"Î$
1 “ (x 1)%Î$
œ _
(d)
x Ä "c
lim
" ’ x"Î$
1 “ (x 1)%Î$
œ _
27. y œ
" x1
28. y œ
" x1
29. y œ
" #x 4
30. y œ
3 x3
31. y œ
x3 x2
32. y œ
2x x1
œ1
" x#
œ#
2 x1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 2.5 Infinite Limits and Vertical Asymptotes 33. y œ
x# x"
35. y œ
x# % x"
œx"
37. y œ
x# 1 x
œx
œx1
" x"
$ x"
" x
39. Here is one possibility.
34. y œ
x# " x1
œx"
36. y œ
x2 " #x %
œ #" x "
38. y œ
x$ 1 x#
œx
# x1
$ #x %
" x#
40. Here is one possibility.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
95
96
Chapter 2 Limits and Continuity
41. Here is one possibility.
42. Here is one possibility.
43. Here is one possibility.
44. Here is one possibility.
45. Here is one possibility.
46. Here is one possibility.
" x#
47. For every real number B 0, we must find a $ 0 such that for all x, 0 kx 0k $ Ê
" x#
Ê
B ! Í " x#
" x#
#
B0 Í x
" B
" ÈB
Í kxk
. Choose $ œ
" ÈB
, then 0 kxk $ Ê kxk
xÄ!
" B.
B ! Í lxl
" B.
Choose $ œ
Then ! kx 0k $ Ê lxl
" B
Ê
" lx l
" lx l
2 (x 3)#
B ! Í
2 (x 3)#
$ œ É B2 , then 0 kx 3k $ Ê
B0 Í 2 (x 3)#
(x 3)# 2
" B
Í (x 3)#
B 0 so that lim
2
# x Ä $ (x 3)
2 B
B. Now,
x Ä ! lx l 2 (x 3)#
Now,
#
B ! Í (x 5)
Ê kx 5k
" ÈB
Ê
" (x 5)#
" B
Í kx 5k
B so that lim
"
" ÈB
# x Ä & (x 5)
. Choose $ œ
œ _.
B.
Í ! kB $k É B2 . Choose
œ _.
50. For every real number B 0, we must find a $ 0 such that for all x, 0 kx (5)k $ Ê 1 (x 5)#
"
B so that lim
49. For every real number B 0, we must find a $ 0 such that for all x, 0 kx 3k $ Ê Now,
" ÈB
B so that lim x"# œ _.
48. For every real number B 0, we must find a $ 0 such that for all x, ! kx 0k $ Ê " lx l
B. Now,
" ÈB
1 (x 5)#
B.
. Then 0 kx (5)k $
œ _.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 2.5 Infinite Limits and Vertical Asymptotes
97
51. (a) We say that f(x) approaches infinity as x approaches x! from the left, and write lim f(x) œ _, if x Ä x!
for every positive number B, there exists a corresponding number $ 0 such that for all x, x! $ x x! Ê f(x) B. (b) We say that f(x) approaches minus infinity as x approaches x! from the right, and write lim f(x) œ _, x Ä x!
if for every positive number B (or negative number B) there exists a corresponding number $ 0 such that for all x, x! x x! $ Ê f(x) B. (c) We say that f(x) approaches minus infinity as x approaches x! from the left, and write lim f(x) œ _, x Ä x!
if for every positive number B (or negative number B) there exists a corresponding number $ 0 such that for all x, x! $ x x! Ê f(x) B. 52. For B 0,
" x
B 0 Í x B" . Choose $ œ B" . Then ! x $ Ê 0 x
53. For B 0,
" x
B 0 Í x" B 0 Í x
Ê B" x Ê 54. For B !,
" x#
" x
B so that lim c xÄ!
" x
" B
Ê
" x#
" x#
Ê
" x#
" 1 x#
" x
œ _.
" B
Í x 2 B" Í x 2 B" . Choose $ œ B" . Then " x#
B 0 so that lim c xÄ#
" x#
œ _.
œ _.
B Í 1 x#
" #B . Then " $ x " Ê " 1 x# B for ! x 1 and " x
B so that lim b xÄ!
B Í ! x 2 B" . Choose $ œ B" . Then # x # $ Ê ! x # $ Ê ! x 2
B ! so that lim b xÄ#
57. y œ sec x
" x
œ _.
B Í x " # B Í (x 2)
56. For B 0 and ! x 1, $
Ê
Í B" x. Choose $ œ B" . Then $ x !
2 $ x 2 Ê $ x 2 ! Ê B" x 2 0 Ê 55. For B 0,
" B
" B
Í (" x)(" x) B" . Now
$ x 1 0 Ê " x $ x near 1 Ê
"
lim # x Ä "c " x
" #B
1x #
1 since x 1. Choose Ê (" x)(" x) B" ˆ 1 # x ‰ B"
œ _.
58. y œ sec x
" x#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
" B
98
Chapter 2 Limits and Continuity
59. y œ tan x
61. y œ
" x#
x È 4 x#
63. y œ x#Î$
60. y œ
" x
62. y œ
" È 4 x#
tan x
64. y œ sin ˆ x# 1 1 ‰
" x"Î$
2.6 CONTINUITY 1. No, discontinuous at x œ 2, not defined at x œ 2 2. No, discontinuous at x œ 3, " œ lim c g(x) Á g(3) œ 1.5 xÄ$
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 2.6 Continuity 3. Continuous on [1ß 3] 4. No, discontinuous at x œ 1, 1.5 œ lim c k(x) Á lim b k(x) œ ! xÄ" xÄ" 5. (a) Yes
(b) Yes,
(c) Yes
(d) Yes
6. (a) Yes, f(1) œ 1
lim
x Ä "b
f(x) œ 0
(b) Yes, lim f(x) œ 2 xÄ1
(c) No
(d) No
7. (a) No
(b) No
8. ["ß !) (!ß ") ("ß #) (#ß $) 9. f(2) œ 0, since lim c f(x) œ 2(2) 4 œ 0 œ lim b f(x) xÄ# xÄ# 10. f(1) should be changed to 2 œ lim f(x) xÄ1
11. Nonremovable discontinuity at x œ 1 because lim f(x) fails to exist ( limc f(x) œ 1 and lim b f(x) œ 0). xÄ" xÄ1 xÄ"
Removable discontinuity at x œ 0 by assigning the number lim f(x) œ 0 to be the value of f(0) rather than xÄ!
f(0) œ 1.
12. Nonremovable discontinuity at x œ 1 because lim f(x) fails to exist ( lim c f(x) œ 2 and lim b f(x) œ 1). xÄ" xÄ1 xÄ" Removable discontinuity at x œ 2 by assigning the number lim f(x) œ 1 to be the value of f(2) rather than xÄ#
f(2) œ 2. 13. Discontinuous only when x 2 œ 0 Ê x œ 2
14. Discontinuous only when (x 2)# œ 0 Ê x œ 2
15. Discontinuous only when x# %x $ œ ! Ê (x 3)(x 1) œ 0 Ê x œ 3 or x œ 1 16. Discontinuous only when x# 3x 10 œ 0 Ê (x 5)(x 2) œ 0 Ê x œ 5 or x œ 2 17. Continuous everywhere. ( kx 1k sin x defined for all x; limits exist and are equal to function values.) 18. Continuous everywhere. ( kxk " Á 0 for all x; limits exist and are equal to function values.) 19. Discontinuous only at x œ 0 20. Discontinuous at odd integer multiples of 1# , i.e., x = (2n ") 1# , n an integer, but continuous at all other x. 21. Discontinuous when 2x is an integer multiple of 1, i.e., 2x œ n1, n an integer Ê x œ
n1 # ,
n an integer, but
continuous at all other x. 22. Discontinuous when
1x #
is an odd integer multiple of 1# , i.e.,
1x #
œ (2n 1) 1# , n an integer Ê x œ 2n 1, n an
integer (i.e., x is an odd integer). Continuous everywhere else.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
99
100
Chapter 2 Limits and Continuity
23. Discontinuous at odd integer multiples of 1# , i.e., x = (2n 1) 1# , n an integer, but continuous at all other x. 24. Continuous everywhere since x% 1 1 and " Ÿ sin x Ÿ 1 Ê 0 Ÿ sin# x Ÿ 1 Ê 1 sin# x 1; limits exist and are equal to the function values. 25. Discontinuous when 2x 3 0 or x 3# Ê continuous on the interval 3# ß _‰ . 26. Discontinuous when 3x 1 0 or x
" 3
Ê continuous on the interval 3" ß _‰ .
27. Continuous everywhere: (2x 1)"Î$ is defined for all x; limits exist and are equal to function values. 28. Continuous everywhere: (2 x)"Î& is defined for all x; limits exist and are equal to function values. 29. xlim sin (x sin x) œ sin (1 sin 1) œ sin (1 0) œ sin 1 œ 0, and function continuous at x œ 1. Ä1 30. lim sin ˆ 1# cos (tan t)‰ œ sin ˆ 1# cos (tan (0))‰ œ sin ˆ 1# cos (0)‰ œ sin ˆ 1# ‰ œ 1, and function continuous at t œ !. tÄ!
31. lim sec ay sec# y tan# y 1b œ lim sec ay sec# y sec# yb œ lim sec a(y 1) sec# yb œ sec a(" ") sec# 1b yÄ1
yÄ1
yÄ1
œ sec 0 œ 1, , and function continuous at y œ ". 32. lim tan 14 cos ˆsin x"Î$ ‰‘ œ tan 14 cos (sin(0))‘ œ tan ˆ 14 cos (0)‰ œ tan ˆ 14 ‰ œ 1, and function continuous at x œ !. xÄ!
33. lim cos ’ È19 13 sec 2t “ œ cos ’ È19 13 sec 0 “ œ cos tÄ!
1 È16
œ cos
1 4
œ
È2 # ,
and function continuous at t œ !.
34. lim1 Écsc# x 5È3 tan x œ Écsc# ˆ 16 ‰ 5È3 tan ˆ 16 ‰ œ Ê4 5È3 Š È"3 ‹ œ È9 œ 3, and function continuous at xÄ
'
x œ 1' . 35. g(x) œ
x# 9 x3
(x 3)(x 3) (x 3)
œ
36. h(t) œ
t# 3t 10 t#
37. f(s) œ
s$ " s# 1
38. g(x) œ
œ
œ
œ x 3, x Á 3 Ê g(3) œ lim (x 3) œ 6
(t 5)(t 2) t#
as# s 1b (s 1) (s 1)(s 1)
x# 16 x# 3x 4
œ
xÄ$
œ t 5, t Á # Ê h(2) œ lim (t 5) œ 7 tÄ#
œ
(x 4)(x 4) (x 4)(x 1)
s# s " s1 ,
œ
x4 x1
s Á 1 Ê f(1) œ lim Š s sÄ1
#
s1 s1 ‹
4‰ , x Á 4 Ê g(4) œ lim ˆ xx 1 œ
xÄ%
œ
3 #
8 5
39. As defined, lim c f(x) œ (3)# 1 œ 8 and lim b (2a)(3) œ 6a. For f(x) to be continuous we must have xÄ$ xÄ$ 6a œ 8 Ê a œ 43 .
40. As defined,
lim
x Ä #c
g(x) œ 2 and
4b œ 2 Ê b œ "# .
lim
x Ä #b
g(x) œ b(2)# œ 4b. For g(x) to be continuous we must have
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 2.6 Continuity 41. The function can be extended: f(0) ¸ 2.3.
42. The function cannot be extended to be continuous at x œ 0. If f(0) ¸ 2.3, it will be continuous from the right. Or if f(0) ¸ 2.3, it will be continuous from the left.
43. The function cannot be extended to be continuous at x œ 0. If f(0) œ 1, it will be continuous from the right. Or if f(0) œ 1, it will be continuous from the left.
44. The function can be extended: f(0) ¸ 7.39.
101
45. f(x) is continuous on [!ß "] and f(0) 0, f(1) 0 Ê by the Intermediate Value Theorem f(x) takes on every value between f(0) and f(1) Ê the equation f(x) œ 0 has at least one solution between x œ 0 and x œ 1.
46. cos x œ x Ê (cos x) x œ 0. If x œ 1# , cos ˆ 1# ‰ ˆ 1# ‰ 0. If x œ 1# , cos ˆ 1# ‰ for some x between
1 #
and
1 #
1 #
0. Thus cos x x œ 0
according to the Intermediate Value Theorem.
47. Let f(x) œ x$ 15x 1 which is continuous on [4ß 4]. Then f(4) œ 3, f(1) œ 15, f(1) œ 13, and f(4) œ 5. By the Intermediate Value Theorem, f(x) œ 0 for some x in each of the intervals % x 1, " x 1, and " x 4. That is, x$ 15x 1 œ 0 has three solutions in [%ß 4]. Since a polynomial of degree 3 can have at most 3 solutions, these are the only solutions. 48. Without loss of generality, assume that a b. Then F(x) œ (x a)# (x b)# x is continuous for all values of x, so it is continuous on the interval [aß b]. Moreover F(a) œ a and F(b) œ b. By the Intermediate Value Theorem, since a a # b b, there is a number c between a and b such that F(x) œ a # b .
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102
Chapter 2 Limits and Continuity
49. Answers may vary. Note that f is continuous for every value of x. (a) f(0) œ 10, f(1) œ 1$ 8(1) 10 œ 3. Since $ 1 10, by the Intermediate Value Theorem, there exists a c so that ! c 1 and f(c) œ 1. (b) f(0) œ 10, f(4) œ (4)$ 8(4) 10 œ 22. Since 22 È3 10, by the Intermediate Value Theorem, there exists a c so that 4 c 0 and f(c) œ È3. (c) f(0) œ 10, f(1000) œ (1000)$ 8(1000) 10 œ 999,992,010. Since 10 5,000,000 999,992,010, by the Intermediate Value Theorem, there exists a c so that ! c 1000 and f(c) œ 5,000,000. 50. All five statements ask for the same information because of the intermediate value property of continuous functions. (a) A root of f(x) œ x$ 3x 1 is a point c where f(c) œ 0. (b) The points where y œ x$ crosses y œ 3x 1 have the same y-coordinate, or y œ x$ œ 3x 1 Ê f(x) œ x$ 3x 1 œ 0. (c) x$ 3x œ 1 Ê x$ 3x 1 œ 0. The solutions to the equation are the roots of f(x) œ x$ 3x 1. (d) The points where y œ x$ 3x crosses y œ 1 have common y-coordinates, or y œ x$ 3x œ 1 Ê f(x) œ x$ 3x 1 œ !. (e) The solutions of x$ 3x 1 œ 0 are those points where f(x) œ x$ 3x 1 has value 0. 51. Answers may vary. For example, f(x) œ
sin (x 2) x2
is discontinuous at x œ 2 because it is not defined there.
However, the discontinuity can be removed because f has a limit (namely 1) as x Ä 2. 52. Answers may vary. For example, g(x) œ
" x1
has a discontinuity at x œ 1 because lim g(x) does not exist. x Ä "
Š lim c g(x) œ _ and lim b g(x) œ _.‹ x Ä " x Ä " 53. (a) Suppose x! is rational Ê f(x! ) œ 1. Choose % œ "# . For any $ 0 there is an irrational number x (actually infinitely many) in the interval (x! $ ß x! $ ) Ê f(x) œ 0. Then 0 kx x! k $ but kf(x) f(x! )k œ 1 "# œ %, so x lim f(x) fails to exist Ê f is discontinuous at x! rational. Äx !
On the other hand, x! irrational Ê f(x! ) œ 0 and there is a rational number x in (x! $ ß x! $ ) Ê f(x) œ 1. Again x lim f(x) fails to exist Ê f is discontinuous at x! irrational. That is, f is discontinuous at Äx !
every point. (b) f is neither right-continuous nor left-continuous at any point x! because in every interval (x! $ ß x! ) or (x! ß x! $ ) there exist both rational and irrational real numbers. Thus neither limits lim f(x) and x Ä x!
lim f(x) exist by the same arguments used in part (a).
x Ä x!
54. Yes. Both f(x) œ x and g(x) œ x g ˆ "# ‰ œ 0 Ê
f(x) g(x)
" #
are continuous on [!ß "]. However
f(x) g(x)
is undefined at x œ
" #
since
is discontinuous at x œ "# .
55. No. For instance, if f(x) œ 0, g(x) œ ÜxÝ, then h(x) œ 0 aÜxÝb œ 0 is continuous at x œ 0 and g(x) is not. 56. Let f(x) œ œ
" x1
" (x 1) 1
œ
and g(x) œ x 1. Both functions are continuous at x œ 0. The composition f ‰ g œ f(g(x)) " x
is discontinuous at x œ 0, since it is not defined there. Theorem 10 requires that f(x) be
continuous at g(0), which is not the case here since g(0) œ 1 and f is undefined at 1. 57. Yes, because of the Intermediate Value Theorem. If f(a) and f(b) did have different signs then f would have to equal zero at some point between a and b since f is continuous on [aß b].
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 2.7 Tangents and Derivatives 58. Let f(x) be the new position of point x and let d(x) œ f(x) x. The displacement function d is negative if x is the left-hand point of the rubber band and positive if x is the right-hand point of the rubber band. By the Intermediate Value Theorem, d(x) œ 0 for some point in between. That is, f(x) œ x for some point x, which is then in its original position. 59. If f(0) œ 0 or f(1) œ 1, we are done (i.e., c œ 0 or c œ 1 in those cases). Then let f(0) œ a 0 and f(1) œ b 1 because 0 Ÿ f(x) Ÿ 1. Define g(x) œ f(x) x Ê g is continuous on [0ß 1]. Moreover, g(0) œ f(0) 0 œ a 0 and g(1) œ f(1) 1 œ b 1 0 Ê by the Intermediate Value Theorem there is a number c in (!ß ") such that g(c) œ 0 Ê f(c) c œ 0 or f(c) œ c. 60. Let % œ
kf(c)k #
0. Since f is continuous at x œ c there is a $ 0 such that kx ck $ Ê kf(x) f(c)k %
Ê f(c) % f(x) f(c) %. If f(c) 0, then % œ "# f(c) Ê " #
" #
If f(c) 0, then % œ f(c) Ê
f(c) f(x) 3 #
3 #
f(c) f(x)
f(c) Ê f(x) 0 on the interval (c $ ß c $ ). " #
f(c) Ê f(x) 0 on the interval (c $ ß c $ ).
61. By Exercises 52 in Section 2.3, we have xlim faxb œ L Í lim fac hb œ L. Äc hÄ0
Thus, faxb is continuous at x œ c Í xlim faxb œ facb Í lim fac hb œ facb. Äc hÄ0
62. By Exercise 61, it suffices to show that lim sinac hb œ sin c and lim cosac hb œ cos c. hÄ0
hÄ0
Now lim sinac hb œ lim asin cbacos hb acos cbasin hb‘ œ asin cbŠ lim cos h‹ acos cbŠ lim sin h‹ hÄ0
hÄ0
hÄ0
hÄ0
By Example 6 Section 2.2, lim cos h œ " and lim sin h œ !. So lim sinac hb œ sin c and thus faxb œ sin x is hÄ0
continuous at x œ c. Similarly,
hÄ0
hÄ0
lim cosac hb œ lim acos cbacos hb asin cbasin hb‘ œ acos cbŠ lim cos h‹ asin cbŠ lim sin h‹ œ cos c.
hÄ0
hÄ0
Thus, gaxb œ cos x is continuous at x œ c.
hÄ0
63. x ¸ 1.8794, 1.5321, 0.3473
64. x ¸ 1.4516, 0.8547, 0.4030
65. x ¸ 1.7549
66. x ¸ 1.5596
67. x ¸ 3.5156
68. x ¸ 3.9058, 3.8392, 0.0667
69. x ¸ 0.7391
70. x ¸ 1.8955, 0, 1.8955
hÄ0
2.7 TANGENTS AND DERIVATIVES 1. P" : m" œ 1, P# : m# œ 5
2. P" : m" œ 2, P# : m# œ 0
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Chapter 2 Limits and Continuity
3. P" : m" œ 5# , P# : m# œ "# 5. m œ lim
hÄ!
4. P" : m" œ 3, P# : m# œ 3
c4 (" h)# d a4 (1)# b h
a1 2h h# b1 h hÄ!
œ lim
œ lim
hÄ!
h(# h) h
œ 2;
at ("ß $): y œ $ #(x (1)) Ê y œ 2x 5, tangent line
6. m œ lim
hÄ!
c(1 h 1)# 1d c(" ")# 1d h
h#
œ lim
hÄ! h
œ lim h œ 0; at ("ß "): y œ 1 0(x 1) Ê y œ 1, hÄ!
tangent line
È 2È 1 h 2È 1 œ lim 2 1 h h 2 h hÄ! hÄ! 4(1 h) 4 œ lim œ lim È1 2h 1 h Ä ! 2h ŠÈ1 h 1‹ hÄ!
7. m œ lim
†
2È 1 h 2 2È 1 h #
œ 1;
at ("ß #): y œ 2 1(x 1) Ê y œ x 1, tangent line
8. m œ lim
hÄ!
"
(1 h)#
("" )#
h
a2h h# b # h Ä ! h(1 h)
œ lim
1 (1 h)# # h Ä ! h(1h) 2h lim # œ 2; h Ä ! (1 h)
œ lim œ
at ("ß "): y œ 1 2(x (1)) Ê y œ 2x 3, tangent line
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 2.7 Tangents and Derivatives (2 h)$ (2)$ h
9. m œ lim
hÄ!
8 12h 6h# h$ 8 h
œ lim
hÄ!
œ lim a12 6h h# b œ 12; hÄ!
at (2ß 8): y œ 8 12(x (2)) Ê y œ 12x 16, tangent line
"
(# h)$
10. m œ lim
h
hÄ!
œ
"2 8(8)
hÄ!
hÄ!
at ˆ#ß "8 ‰ : y œ 8" Ê yœ
11. m œ lim
hÄ!
x
" #,
12 6h h# 8(2 h)$
œ lim
3 œ 16 ;
3 16
8 (# h)$ 8h(# h)$
œ lim
a12h 6h# h$ b 8h(# h)$
œ lim
hÄ!
(#" )$
3 16 (x
(2))
tangent line
c(2 h)# 1d 5 h
œ lim
hÄ!
a5 4h h# b 5 h
hÄ!
at (2ß 5): y 5 œ 4(x 2), tangent line 12. m œ lim
hÄ!
c(" h) 2(1 h)# d (1) h
œ lim
hÄ!
h(4 h) h
œ lim
a1 h 2 4h 2h# b 1 h
hÄ!
3h (3 h) 2
3
h
œ lim
hÄ!
(3 h) 3(h 1) h(h 1)
h Ä ! h(h 1)
at ($ß $): y 3 œ 2(x 3), tangent line 14. m œ lim
hÄ!
8 (2 h)#
2
h
hÄ!
hÄ!
(2 h)$ 8 h
œ lim
hÄ!
œ lim
a8 12h 6h# h$ b 8 h
hÄ!
at (2ß )): y 8 œ 12(t 2), tangent line 16. m œ lim
hÄ!
c(1 h)$ 3(1 h)d 4 h
a1 3h 3h# h$ 3 3hb 4 h
œ lim
hÄ!
at ("ß %): y 4 œ 6(t 1), tangent line È4 h 2 h hÄ!
17. m œ lim
È4 h 2 h hÄ!
œ lim
œ "4 ; at (%ß #): y 2 œ 18. m œ lim
hÄ!
œ
" È9 3
È(8 h) 1 3 h
" 4
†
È4 h 2 È4 h 2
œ lim
hÄ!
h a12 6h h# b h
œ lim
œ 3;
œ 2;
8 2 a4 4h h# b h(2 h)# hÄ!
8 2(2 h)# # h Ä ! h(2 h)
œ lim
at (2ß 2): y 2 œ 2(x 2) 15. m œ lim
2h
œ lim
h(3 2h) h
œ lim
at ("ß "): y 1 œ 3(x 1), tangent line 13. m œ lim
œ %;
œ lim
2h(4 h) h(2 h)#
œ
8 4
œ 2;
œ 12;
œ lim
hÄ!
(4 h) 4
h Ä ! h ŠÈ4 h #‹
h a6 3h h# b h
œ lim
œ 6;
h
h Ä ! h ŠÈ4 h #‹
œ
" È4 #
(x 4), tangent line
œ lim
hÄ!
È9 h 3 h
œ 6" ; at (8ß 3): y 3 œ
19. At x œ 1, y œ 5 Ê m œ lim
hÄ!
" 6
†
È9 h 3 È9 h 3
œ lim
(9 h) 9
h Ä ! h ŠÈ9 h 3‹
œ lim
h
h Ä ! h ŠÈ9 h 3‹
(x 8), tangent line
5(" h)# 5 h
œ lim
hÄ!
5 a1 2h h# b 5 h
œ lim
hÄ!
5h(2 h) h
œ 10, slope
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Chapter 2 Limits and Continuity c1 (2 h)# d (3) h
20. At x œ 2, y œ 3 Ê m œ lim
hÄ!
21. At x œ 3, y œ
" #
"
(3 h) 1
Ê m œ lim
#"
22. At x œ 0, y œ 1 Ê m œ lim
hÄ!
h1 h1
hÄ!
hÄ!
(1) h
a1 4 4h h# b 3 h
2 (2 h) 2h(2 h)
œ lim
h
hÄ!
œ lim
hÄ!
hÄ!
h
œ lim
h Ä ! 2h(2 h)
(h 1) (h ") h(h 1)
œ lim
œ lim
œ lim
h(4 h) h
œ 4, slope
œ "4 , slope 2h
h Ä ! h(h 1)
œ 2, slope
c(x h)# 4(x h) 1d ax# 4x 1b h hÄ! a2xh h# 4hb lim œ lim (2x h 4) œ 2x h hÄ! hÄ!
23. At a horizontal tangent the slope m œ 0 Ê 0 œ m œ lim ax# 2xh h# 4x 4h 1b ax# 4x 1b h hÄ!
œ lim
œ
4;
2x 4 œ 0 Ê x œ 2. Then f(2) œ 4 8 1 œ 5 Ê (2ß 5) is the point on the graph where there is a horizontal tangent. 24. 0 œ m œ lim
hÄ!
c(x h)$ 3(x h)d ax$ 3xb h
3x# h 3xh# h$ 3h h
œ lim
hÄ!
œ lim
hÄ!
ax$ 3x# h 3xh# h$ 3x 3hb ax$ 3xb h
œ lim a3x# 3xh h# 3b œ 3x# 3; 3x# 3 œ 0 Ê x œ 1 or x œ 1. Then hÄ!
f(1) œ 2 and f(1) œ 2 Ê ("ß 2) and ("ß 2) are the points on the graph where a horizontal tangent exists. "
(x h) 1
25. 1 œ m œ lim
x " 1
h
hÄ!
(x 1) (x h 1) h(x 1)(x h 1)
œ lim
hÄ!
h
œ lim
h Ä ! h(x 1)(x h 1)
œ (x " 1)#
Ê (x 1)# œ 1 Ê x# 2x œ 0 Ê x(x 2) œ 0 Ê x œ 0 or x œ 2. If x œ 0, then y œ 1 and m œ 1 Ê y œ 1 (x 0) œ (x 1). If x œ 2, then y œ 1 and m œ 1 Ê y œ 1 (x 2) œ (x 3). 26.
" 4
œ m œ lim
Èx h Èx
œ lim
h
y œ 2 "4 (x 4) œ
hÄ!
f(2 h) f(2) h
x 4
Èx h Èx h
hÄ!
h Ä ! h ŠÈx h Èx‹
27. lim
œ lim
h
hÄ!
œ
" #È x
. Thus,
" 4
œ
†
Èx h Èx Èx h Èx
" #Èx
(x h) x
œ lim
h Ä ! h ŠÈx h Èx‹
Ê Èx œ 2 Ê x œ 4 Ê y œ 2. The tangent line is
1.
œ lim
hÄ!
a100 4.9(# h)# b a100 4.9(2)# b h
4.9 a4 4h h# b 4.9(4) h
œ lim
hÄ!
œ lim (19.6 4.9h) œ 19.6. The minus sign indicates the object is falling downward at a speed of hÄ!
19.6 m/sec. f(10 h) f(10) h hÄ!
28. lim
3(10 h)# 3(10)# h hÄ!
œ lim
29. lim
f(3 h) f(3) h
œ lim
30. lim
f(2 h) f(2) h
œ lim
hÄ!
hÄ!
hÄ!
hÄ!
1(3 h)# 1(3)# h 41 3
œ lim
(2 h)$ 431 (2)$ h
f(0 h) f(0) h hÄ!
31. Slope at origin œ lim
3 a20h h# b h hÄ!
œ lim
hÄ!
œ lim
1 c9 6h h# 9d h 41 3
hÄ!
h# sin ˆ "h ‰ h hÄ!
œ lim
œ 60 ft/sec. œ lim 1(6 h) œ 61
c12h 6h# h$ d h
hÄ!
œ lim
hÄ!
41 3
c12 6h h# d œ 161
œ lim h sin ˆ h" ‰ œ 0 Ê yes, f(x) does have a tangent at hÄ!
the origin with slope 0. 32. lim
hÄ!
g(0 h) g(0) h
œ lim
hÄ!
h sin ˆ "h ‰ h
œ lim sin h" . Since lim sin hÄ!
hÄ!
" h
does not exist, f(x) has no tangent at
the origin.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 2.7 Tangents and Derivatives 33.
lim
h Ä !c
lim
hÄ!
34.
f(0 h) f(0) h f(0 h) f(0) h
œ lim c hÄ!
1 0 h
œ _, and lim b hÄ!
f(0 h) f(0) h
10 h
œ lim b hÄ!
œ _ Ê yes, the graph of f has a vertical tangent at the origin.
œ _, and lim b U(0 h)h U(0) œ lim b hÄ! hÄ! does not have a vertical tangent at (!ß ") because the limit does not exist. lim
h Ä !c
œ _. Therefore,
U(0 h) U(0) h
œ lim c hÄ!
01 h
11 h
œ 0 Ê no, the graph of f
35. (a) The graph appears to have a cusp at x œ 0.
(b)
lim
h Ä !c
f(0 h) f(0) h
œ lim c hÄ!
h#Î& 0 h
œ lim c hÄ!
" h$Î&
œ _ and lim b hÄ!
" h$Î&
œ _ Ê limit does not exist
Ê the graph of y œ x#Î& does not have a vertical tangent at x œ 0.
36. (a) The graph appears to have a cusp at x œ 0.
(b)
lim
h Ä !c
f(0 h) f(0) h
œ lim c hÄ!
h%Î& 0 h
œ lim c hÄ!
" h"Î&
œ _ and lim b hÄ!
" h"Î&
œ _ Ê limit does not exist
Ê y œ x%Î& does not have a vertical tangent at x œ 0.
37. (a) The graph appears to have a vertical tangent at x œ !.
(b)
f(0 h) f(0) h hÄ!
lim
h"Î& 0 h hÄ!
œ lim
œ lim
"
%Î& hÄ! h
œ _ Ê y œ x"Î& has a vertical tangent at x œ 0.
38. (a) The graph appears to have a vertical tangent at x œ 0.
(b)
lim
hÄ!
f(0 h) f(0) h
at x œ 0.
œ lim
hÄ!
h$Î& 0 h
œ lim
"
#Î& hÄ! h
œ _ Ê the graph of y œ x$Î& has a vertical tangent
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Chapter 2 Limits and Continuity
39. (a) The graph appears to have a cusp at x œ 0.
(b)
lim
h Ä !c
f(0 h) f(0) h
œ lim c hÄ!
4h#Î& 2h h
œ lim c hÄ!
4 h$Î&
2 œ _ and lim b hÄ!
4 h$Î&
#œ_
Ê limit does not exist Ê the graph of y œ 4x#Î& 2x does not have a vertical tangent at x œ 0.
40. (a) The graph appears to have a cusp at x œ 0.
(b)
lim
hÄ!
f(0 h) f(0) h
œ lim
hÄ!
h&Î$ 5h#Î$ h
œ lim h#Î$ hÄ!
5 h"Î$
œ 0 lim
5
"Î$ hÄ! h
y œ x&Î$ 5x#Î$ does not have a vertical tangent at x œ !.
does not exist Ê the graph of
41. (a) The graph appears to have a vertical tangent at x œ 1 and a cusp at x œ 0.
(b) x œ 1:
(1 h)#Î$ (1 h 1)"Î$ " h hÄ! #Î$ "Î$
lim
Ê yœx
x œ 0:
(x 1)
lim f(0 h)h f(0) hÄ!
(1 h)#Î$ h"Î$ " h hÄ!
œ lim
has a vertical tangent at x œ 1;
h#Î$ (h 1)"Î$ (1)"Î$ h hÄ! #Î$ "Î$
œ lim
does not exist Ê y œ x
œ _
(x 1)
" œ lim ’ h"Î$
hÄ!
(h ")"Î$ h
h" “
does not have a vertical tangent at x œ 0.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 2.7 Tangents and Derivatives 42. (a) The graph appears to have vertical tangents at x œ 0 and x œ 1.
(b) x œ 0:
h"Î$ (h 1)"Î$ (")"Î$ h hÄ!
f(0 h) f(0) h hÄ!
œ lim
f(1 h) f(1) h
œ lim
lim
œ _ Ê y œ x"Î$ (x 1)"Î$ has a
vertical tangent at x œ 0;
x œ 1:
lim
hÄ!
hÄ!
(1 h)"Î$ (" h 1)"Î$ 1 h
œ _ Ê y œ x"Î$ (x 1)"Î$ has a
vertical tangent at x œ ".
43. (a) The graph appears to have a vertical tangent at x œ 0.
(b)
lim
h Ä !b
f(0 h) f(0) h
œ lim b xÄ!
Èh 0 h
œ lim
È kh k 0
f(0 h) f(0) h
"
h Ä ! Èh
œ lim c œ lim c h hÄ! hÄ! Ê y has a vertical tangent at x œ 0. lim
h Ä !c
œ _; È kh k kh k
œ lim c hÄ!
" È kh k
œ_
44. (a) The graph appears to have a cusp at x œ 4.
(b)
lim
f(4 h) f(4) h
œ lim b hÄ!
Èk4 (4 h)k 0 h
lim
f(4 h) f(4) h
œ lim c hÄ!
Èk4 (4 h)k h
h Ä !b h Ä !c
œ lim b hÄ!
œ lim c hÄ!
È kh k h
È kh k lhl
œ lim b hÄ!
œ lim c hÄ!
" Èh
" È kh k
œ _;
œ _
Ê y œ È% x does not have a vertical tangent at x œ 4. 45-48. Example CAS commands: Maple: f := x -> x^3 + 2*x;x0 := 0; plot( f(x), x=x0-1/2..x0+3, color=black, # part (a) title="Section 2.7, #45(a)" ); q := unapply( (f(x0+h)-f(x0))/h, h ); # part (b) L := limit( q(h), h=0 ); # part (c) sec_lines := seq( f(x0)+q(h)*(x-x0), h=1..3 ); # part (d) tan_line := f(x0) + L*(x-x0); plot( [f(x),tan_line,sec_lines], x=x0-1/2..x0+3, color=black, Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Chapter 2 Limits and Continuity
linestyle=[1,2,5,6,7], title="Section 2.7, #45(d)", legend=["y=f(x)","Tangent line at x=0","Secant line (h=1)", "Secant line (h=2)","Secant line (h=3)"] ); Mathematica: (function and value for x0 may change) Clear[f, m, x, h] x0 œ p; f[x_]: œ Cos[x] 4Sin[2x] Plot[f[x], {x, x0 1, x0 3}] dq[h_]: œ (f[x0+h] f[x0])/h m œ Limit[dq[h], h Ä 0] ytan: œ f[x0] m(x x0) y1: œ f[x0] dq[1](x x0) y2: œ f[x0] dq[2](x x0) y3: œ f[x0] dq[3](x x0) Plot[{f[x], ytan, y1, y2, y3}, {x, x0 1, x0 3}] CHAPTER 2 PRACTICE EXERCISES 1. At x œ 1: Ê
lim
x Ä "c
f(x) œ
lim
x Ä "b
f(x) œ 1
lim f(x) œ 1 œ f(1)
x Ä 1
Ê f is continuous at x œ 1. At x œ 0: lim c f(x) œ lim b f(x) œ 0 Ê lim f(x) œ 0. xÄ!
xÄ!
xÄ!
But f(0) œ 1 Á lim f(x) xÄ!
Ê f is discontinuous at x œ 0. If we define fa!b œ !, then the discontinuity at x œ ! is removable. At x œ 1: lim c f(x) œ 1 and lim b f(x) œ 1 xÄ"
Ê lim f(x) does not exist
xÄ"
xÄ1
Ê f is discontinuous at x œ 1. 2. At x œ 1: Ê
lim
x Ä "c
f(x) œ 0 and
lim
x Ä "b
f(x) œ 1
lim f(x) does not exist
x Ä "
Ê f is discontinuous at x œ 1. At x œ 0: lim c f(x) œ _ and lim b f(x) œ _ xÄ!
Ê lim f(x) does not exist
xÄ!
xÄ!
Ê f is discontinuous at x œ 0. At x œ 1: lim c f(x) œ lim b f(x) œ 1 Ê lim f(x) œ 1. xÄ"
xÄ1
xÄ"
But f(1) œ 0 Á lim f(x) xÄ1
Ê f is discontinuous at x œ 1. If we define fa"b œ ", then the discontinuity at x œ " is removable. 3. (a) (b)
lim a3fatbb œ 3 lim fatb œ 3(7) œ 21
t Ä t!
t Ä t!
#
#
lim afatbb œ Š lim fatb‹ œ a(b# œ 49
t Ä t!
t Ä t!
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 2 Practice Exercises (c) (d) (e) (f)
111
lim afatb † gatbb œ lim fatb † lim gatb œ (7)(0) œ 0
t Ä t!
t Ä t!
lim fatb t Ä t! g(t)7
Ät
t Ä t!
lim fatb
œ
Ät
t
t
Ät
t
!
Ät
t
!
7 07
œ
lim gatb lim 7
t
!
Ät
lim fatb
œ
!
lim agatb 7b
!
œ1
lim cos agatbb œ cos Š lim gatb‹ œ cos ! œ 1
t Ä t!
t Ä t!
lim kfatbk œ ¹ lim fatb¹ œ k7k œ 7
t Ä t!
t Ä t!
(g) lim afatb gatbb œ lim fatb lim gatb œ 7 0 œ 7 t Ä t!
(h)
4. (a) (b) (c) (d) (e) (f)
t Ä t!
lim Š " ‹ t Ä t! fatb
œ
" lim fatb
t
Ät
t Ä t!
" 7
œ
!
œ 71
lim g(x) œ lim g(x) œ È2
xÄ!
xÄ!
lim ag(x) † f(x)b œ lim g(x) † lim f(x) œ ŠÈ2‹ ˆ "# ‰ œ
xÄ!
xÄ!
xÄ!
lim af(x) g(x)b œ lim f(x) lim g(x) œ
xÄ!
"
lim x Ä ! f(x)
œ
xÄ!
" lim f(x)
œ
xÄ!
" " #
xÄ!
œ2
" #
lim ax f(x)b œ lim x lim f(x) œ 0
xÄ!
xÄ!
f(x)†cos x x 1 xÄ!
lim
xÄ!
lim f(x)† lim cos x
œ
xÄ!
xÄ!
lim x lim 1
xÄ!
xÄ!
œ
ˆ "# ‰ (1) 01
" #
È2 #
È2
œ
" #
œ #"
5. Since lim x œ 0 we must have that lim (4 g(x)) œ 0. Otherwise, if lim (% g(x)) is a finite positive xÄ!
xÄ!
xÄ!
’ 4xg(x) “
’ 4xg(x) “
œ _ and lim b œ _ so the limit could not equal 1 as xÄ! x Ä 0. Similar reasoning holds if lim (4 g(x)) is a finite negative number. We conclude that lim g(x) œ 4. number, we would have lim c xÄ!
xÄ!
6. 2 œ lim
x Ä %
xÄ!
’x lim g(x)“ œ lim x † lim xÄ!
x Ä %
x Ä %
’ lim g(x)“ œ 4 lim xÄ!
(since lim g(x) is a constant) Ê lim g(x) œ xÄ!
xÄ!
2 %
x Ä %
œ #" .
’ lim g(x)“ œ 4 lim g(x) xÄ!
xÄ!
7. (a) xlim faxb œ xlim x"Î$ œ c"Î$ œ facb for every real number c Ê f is continuous on a_ß _b. Äc Äc (b) xlim gaxb œ xlim x$Î% œ c$Î% œ gacb for every nonnegative real number c Ê g is continuous on Ò!ß _Ñ. Äc Äc (c) xlim haxb œ xlim x#Î$ œ Äc Äc (d) xlim kaxb œ xlim x"Î' œ Äc Äc
" c#Î$ " c"Î'
œ hacb for every nonzero real number c Ê h is continuous on a_ß !b and a_ß _b. œ kacb for every positive real number c Ê k is continuous on a!ß _b
8. (a) - ˆˆn "# ‰1ß ˆn "# ‰1‰, where I œ the set of all integers. n−I (b) - an1ß an 1b1b, where I œ the set of all integers. n−I (c) a_ß 1b a1ß _b (d) a_ß !b a!ß _b 9.
(a)
(b) 10. (a)
(x 2)(x 2) x# 4x 4 $ 5x# 14x œ lim x xÄ! x Ä ! x(x 7)(x 2) x2 x2 lim œ _ and lim b x(x 7) x Ä !c x(x 7)
œ lim
lim (x 2)(x 2) x Ä # x(x 7)(x #)
œ lim
lim
#
x 4x 4
xÄ!
lim $ # x Ä # x 5x 14x x# x
lim & % $ x Ä ! x 2x x Now lim c xÄ!
œ
œ lim
x(x 1)
$ # x Ä ! x ax 2x 1b
1 x# (x 1)
œ _ and lim b xÄ!
x2
, x Á 2; the limit does not exist because
x2
, x Á 2, and lim
x Ä ! x(x 7)
œ _
x Ä # x(x 7)
œ lim
x1
# x Ä ! x (x 1)(x 1)
1 x# (x 1)
x2
x Ä # x(x 7)
œ lim
œ _ Ê lim
"
# x Ä 0 x (x 1) #
x x
& % $ x Ä ! x 2x x
œ
0 2(9)
œ0
, x Á 0 and x Á 1.
œ _.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
112
Chapter 2 Limits and Continuity
(b)
x# x
exist because
$ # x Ä " x ax 2x 1b
"
1 Èx 1x
œ lim
12. xlim Äa
x # a# x % a%
œ xlim Äa
13. lim
(x h)# x# h
œ lim
(x h)# x# h xÄ!
œ lim
hÄ!
"
15. lim
#x #
16. lim
(# x)$ 8 x
xÄ!
xÄ!
17.
18.
xÄ!
xÄ!
lim
xÄ1
" x g(x)
3x# 1 g(x)
œ2 Ê
xÄ!
"
x Ä ! 4 #x
xÄ!
xÄ1
5 x#
œ0 Ê
" #
lim 4 g(x) œ 8, since 2$ œ 8. Then lim b g(x) œ 2. xÄ!
x Ä !b
Ê È5 lim
x Ä È5
g(x) œ
" #
Ê
lim
x Ä È5
g(x) œ
%x ) $x $
x Ä #
#! &!
ˆ" œxÄ lim _ $x
% $x#
"
#
#
# &
œ
" x# 24. x lim œ x lim œ Ä _ x # (x " Ä _ " (x x"#
#x $ ##. x Ä lim œxÄ lim _ &x# ( _ & ) ‰ $x$
! "!!
$ x# ( x#
œ
#! &!
œ!!!œ!
œ!
#
%
x (x x( 25. x Ä lim œxÄ lim œ _ _ x 1 _ " "x
$
x x x" #'. x lim œxÄ lim œ_ Ä _ "#x$ "#) _ "# "#) x$
lsin xl lsin xl " 27. x lim Ÿ x lim œ ! since int x Ä _ as x Ä _ Êx lim œ !. Ä _ gx h Ä _ gx h Ä _ gx h
lim
)Ä_
29. x lim Ä_
lcos ) "l )
Ÿ lim
"
l#l
)Ä_ )
x sin x #Èx x sin x
#Î$
È5
lim g(x) œ _ since lim a5 x# b œ 1
# $
28.
" #
xÄ1
x Ä #
#x $ x 21. x lim œ x lim œ Ä _ &x ( Ä _ & (x #
œ2 Ê
œ _ Ê lim g(x) œ 0 since lim a3x# 1b œ 4
lim x Ä # Èg(x)
x 23. x Ä lim _
œ "4
œ lim ax# 6x 12b œ 12
(x g(x)) œ
lim
" #a#
œ lim (2x h) œ h
œ lim
x Ä È&
œ
" #
hÄ!
"Î$
x Ä È&
œ
œ _.
œ lim (2x h) œ 2x
ax$ 6x# 12x 8b 8 x
œ lim
" x # a#
œ xlim Äa
ax# 2hx h# b x# h
2 (2 x) 2x(# x)
œ lim
"
lim [4 g(x)]"Î$ œ 2 Ê ’ lim b 4 g(x)“ x Ä !b xÄ!
19. lim 20.
x
ax # a # b ax # a # b a x # a # b
"
lim # x Ä "b x (x 1)
x Ä 1 1 Èx
, x Á 0 and x Á 1. The limit does not
1
# x Ä " x (x 1)
œ lim
ax# 2hx h# b x# h xÄ!
14. lim
"
" Èx
x Ä 1 ˆ1 È x ‰ ˆ 1 È x ‰
hÄ!
œ lim
œ _ and
lim # x Ä "c x (x 1)
11. lim
xÄ1
x(x 1)
œ lim
lim & % $ x Ä " x 2x x
œ ! Ê lim
œ x lim Ä_
)Ä_
" sinx x È#x " sinx x
&Î$
x x " x 30. x lim œ x lim #x œ Ä _ x#Î$ cos# x Ä _Œ " cos#Î$
lcos ) "l )
œ !.
"!! "!
œ"
œ
"! "!
œ"
x
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
œ
# &
Chapter 2 Practice Exercises 31. At x œ 1: œ
lim
x Ä "c
lim
x Ä "c
lim
x Ä "b
f(x) œ
x ax # 1 b x# 1
f(x) œ
œ
lim
x ax # 1 b kx # 1 k
lim
x Ä "c
lim
x Ä "c
x Ä "b
x œ 1, and
x ax # 1 b kx # 1 k
œ
x ax # 1 b
lim # x Ä "b ax "b
œ lim (x) œ (1) œ 1. Since x Ä 1
lim f(x) Á lim b f(x) x Ä "c x Ä " Ê
lim f(x) does not exist, the function f cannot be
x Ä 1
extended to a continuous function at x œ 1. At x œ 1:
lim f(x) œ lim c xÄ"
x Ä "c
#
x ax # 1 b kx # 1 k
œ lim c xÄ" #
x ax # 1 b ax # 1 b
œ lim c (x) œ 1, and xÄ"
lim f(x) œ lim b xkaxx# 11k b œ lim b x axx# "1b œ lim b x œ 1. Again lim f(x) does not exist so f xÄ1 xÄ" xÄ" xÄ1 cannot be extended to a continuous function at x œ 1 either.
x Ä "b
32. The discontinuity at x œ 0 of f(x) œ sin ˆ "x ‰ is nonremovable because lim sin xÄ!
" x
does not exist.
33. Yes, f does have a continuous extension to a œ 1: " define f(1) œ lim xxÈ œ 43 . % x xÄ1
34. Yes, g does have a continuous extension to a œ 1# : ) 5 g ˆ 1# ‰ œ lim1 45)cos #1 œ 4 . )Ä #
35. From the graph we see that lim h(t) Á lim h(t) tÄ! tÄ! so h cannot be extended to a continuous function at a œ 0.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
113
114
Chapter 2 Limits and Continuity
36. From the graph we see that lim c k(x) Á lim b k(x) xÄ! xÄ! so k cannot be extended to a continuous function at a œ 0.
37. (a) f(1) œ 1 and f(2) œ 5 Ê f has a root between 1 and 2 by the Intermediate Value Theorem. (b), (c) root is 1.32471795724 38. (a) f(2) œ 2 and f(0) œ 2 Ê f has a root between 2 and 0 by the Intermediate Value Theorem. (b), (c) root is 1.76929235424 CHAPTER 2 ADDITIONAL AND ADVANCED EXERCISES 1. (a)
x xx
0.1 0.7943
0.01 0.9550
0.001 0.9931
10
100
1000
0.3679
0.3679
0.3679
0.0001 0.9991
0.00001 0.9999
Apparently, lim b xx œ 1 xÄ! (b)
2. (a)
x ˆ "x ‰"ÎÐln xÑ Apparently,
"ÎÐln xÑ lim ˆ " ‰ xÄ_ x
œ 0.3678 œ
" e
(b)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 2 Additional and Advanced Exercises 115 3.
lim L œ lim c L! É" vc# œ L! É1 vÄcc# œ L! É1 cc# œ 0 vÄc The left-hand limit was needed because the function L is undefined if v c (the rocket cannot move faster than the speed of light). lim v#
#
v Ä cc
#
4. ¹
Èx #
1¹ 0.2 Ê 0.2
Èx #
1 0.2 Ê 0.8
Èx #
1.2 Ê 1.6 Èx 2.4 Ê 2.56 x 5.76.
¹
Èx #
1¹ 0.1 Ê 0.1
Èx #
1 0.1 Ê 0.9
Èx #
1.1 Ê 1.8 Èx 2.2 Ê 3.24 x 4.84.
5. k10 (t 70) ‚ 10% 10k 0.0005 Ê k(t 70) ‚ 10% k 0.0005 Ê 0.0005 (t 70) ‚ 10% 0.0005 Ê 5 t 70 5 Ê 65° t 75° Ê Within 5° F. 6. We want to know in what interval to hold values of h to make V satisfy the inequality lV "!!!l œ l$'1h "!!!l Ÿ "!. To find out, we solve the inequality: **! l$'1h "!!!l Ÿ "! Ê "! Ÿ $'1h "!!! Ÿ "! Ê **! Ÿ $'1h Ÿ "!"! Ê $' 1 Ÿ hŸ
"!"! $'1
Ê )Þ) Ÿ h Ÿ )Þ*. where 8.8 was rounded up, to be safe, and 8.9 was rounded down, to be safe. The interval in which we should hold h is about )Þ* )Þ) œ !Þ" cm wide (1 mm). With stripes 1 mm wide, we can expect to measure a liter of water with an accuracy of 1%, which is more than enough accuracy for cooking.
7. Show lim f(x) œ lim ax# 7b œ ' œ f(1). xÄ1
xÄ1
Step 1: kax# 7b 6k % Ê % x# 1 % Ê 1 % x# 1 % Ê È1 % x È1 %. Step 2: kx 1k $ Ê $ x 1 $ Ê $ " x $ ". Then $ " œ È1 % or $ " œ È1 %. Choose $ œ min š1 È1 %ß È1 % 1› , then 0 kx 1k $ Ê kax# (b 6k % and lim f(x) œ 6. By the continuity test, f(x) is continuous at x œ 1. xÄ1
8. Show lim" g(x) œ lim" xÄ
xÄ
%
" 2x
œ 2 œ g ˆ "4 ‰ .
%
Step 1: ¸ #"x 2¸ % Ê % #"x # % Ê # % #"x # % Ê Step 2: ¸B "4 ¸ $ Ê $ x 4" $ Ê $ 4" x $ 4" . Then $ Choose $ œ
" 4
œ
" 4 #%
% 4(#%)
Ê $œ
" 4
" 4 #%
œ
% 4(2 %)
, or $
" 4
œ
, the smaller of the two values. Then 0 ¸x
By the continuity test, g(x) is continuous at x œ
" 4
" 4 #% Ê 4" ¸ $
" 4#%
x
" 4 #% ¸ #"x
" 4#%
.
" 4
% 4(2 %)
$œ
œ
Ê
2¸ % and lim"
.
xÄ
%
" #x
œ 2.
.
9. Show lim h(x) œ lim È2x 3 œ " œ h(2). xÄ#
xÄ#
Step 1: ¹È2x 3 1¹ % Ê % È2x 3 " % Ê " % È2x 3 " % Ê
(1 %)# $ #
x
(" %)# 3 . #
Step 2: kx 2k $ Ê $ x 2 $ or $ # x $ #. (" % )# $ Ê $œ # (" % Ñ # $ (" %Ñ# " #œ # #
Then $ # œ
#
Ê $œ
œ%
# (" %)# $ œ " (1# %) # # %# . Choose $ œ %
œ%
%# #,
%# #
, or $ # œ
(" %)# $ #
the smaller of the two values . Then,
! kx 2k $ Ê ¹È2x 3 "¹ %, so lim È2x 3 œ 1. By the continuity test, h(x) is continuous at x œ 2. xÄ#
10. Show lim F(x) œ lim È9 x œ # œ F(5). xÄ&
xÄ&
Step 1: ¹È9 x 2¹ % Ê % È9 x # % Ê 9 (2 %)# x * (# %)# . Step 2: 0 kx 5k $ Ê $ x & $ Ê $ & x $ &. Then $ & œ * (# %)# Ê $ œ (# %)# % œ %# #%, or $ & œ * (# %)# Ê $ œ % (# %)# œ %# #%.
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116
Chapter 2 Limits and Continuity
Choose $ œ %# #%, the smaller of the two values. Then, ! kx 5k $ Ê ¹È9 x #¹ %, so lim È9 x œ #. By the continuity test, F(x) is continuous at x œ 5.
xÄ&
11. Suppose L" and L# are two different limits. Without loss of generality assume L# L" . Let % œ
" 3
(L# L" ).
Since x lim f(x) œ L" there is a $" 0 such that 0 kx x! k $" Ê kf(x) L" k % Ê % f(x) L" % Äx !
Ê "3 (L# L" ) L" f(x)
" 3
(L# L" ) L" Ê 4L" L# 3f(x) 2L" L# . Likewise, x lim f(x) œ L# Ä x! so there is a $# such that 0 kx x! k $# Ê kf(x) L# k % Ê % f(x) L# % Ê "3 (L# L" ) L# f(x) 3" (L# L" ) L# Ê 2L# L" 3f(x) 4L# L" Ê L" 4L# 3f(x) 2L# L" . If $ œ min e$" ß $# f both inequalities must hold for 0 kx x! k $ : 4L" L# 3f(x) 2L" L# Ê 5(L" L# ) 0 L" L# . That is, L" L# 0 and L" L# 0, L" %L# 3f(x) 2L# L" a contradiction. 12. Suppose xlim f(x) œ L. If k œ !, then xlim kf(x) œ xlim 0 œ ! œ ! † xlim f(x) and we are done. Äc Äc Äc Äc % If k Á 0, then given any % !, there is a $ ! so that ! lx cl $ Ê lfaxb Ll l5l Ê lkllfaxb Ll % Ê lkafaxb Lb| % Ê lakfaxbb akLbl %. Thus, xlim kf(x) œ kL œ kŠxlim f(x)‹. Äc Äc 13. (a) Since x Ä 0 , 0 x$ x 1 Ê ax$ xb Ä 0 Ê
lim f ax$ xb œ lim c f(y) œ B where y œ x$ x. yÄ!
x Ä !b
(b) Since x Ä 0 , 1 x x$ 0 Ê ax$ xb Ä 0 Ê
(c) Since x Ä 0 , 0 x% x# 1 Ê ax# x% b Ä 0 Ê
lim f ax$ xb œ lim b f(y) œ A where y œ x$ x. yÄ!
x Ä !c
lim f ax# x% b œ lim b f(y) œ A where y œ x# x% . yÄ!
x Ä !b
(d) Since x Ä 0 , 1 x 0 Ê ! x% x# 1 Ê ax# x% b Ä 0 Ê
lim f ax# x% b œ A as in part (c).
x Ä !b
14. (a) True, because if xlim (f(x) g(x)) exists then xlim (f(x) g(x)) xlim f(x) œ xlim [(f(x) g(x)) f(x)] Äa Äa Äa Äa œ xlim g(x) exists, contrary to assumption. Äa
(b) False; for example take f(x) œ
" x
and g(x) œ x" . Then neither lim f(x) nor lim g(x) exists, but
lim (f(x) g(x)) œ lim ˆ "x x" ‰ œ lim 0 œ 0 exists.
xÄ!
xÄ!
xÄ!
xÄ!
xÄ!
(c) True, because g(x) œ kxk is continuous Ê g(f(x)) œ kf(x)k is continuous (it is the composite of continuous functions). 1, x Ÿ 0 Ê f(x) is discontinuous at x œ 0. However kf(x)k œ 1 is (d) False; for example let f(x) œ œ 1, x 0 continuous at x œ 0. 15. Show lim f(x) œ lim x Ä 1
x# "
x Ä 1 x 1
œ lim
x Ä 1
(x 1)(x ") (x 1)
Define the continuous extension of f(x) as F(x) œ œ
œ #, x Á 1.
x# 1 x1 ,
2
x Á " . We now prove the limit of f(x) as x Ä 1 , x œ 1
exists and has the correct value. #
Step 1: ¹ xx 1" (#)¹ % Ê %
(x 1)(x ") (x 1)
# % Ê % (x 1) # %, x Á " Ê % " x % ".
Step 2: kx (1)k $ Ê $ x 1 $ Ê $ " x $ ". Then $ " œ % " Ê $ œ %, or $ " œ % " Ê $ œ %. Choose $ œ %. Then ! kx (1)k $ #
Ê ¹ xx 1" a#b¹ % Ê
lim F(x) œ 2. Since the conditions of the continuity test are met by F(x), then f(x) has a
x Ä 1
continuous extension to F(x) at x œ 1.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 2 Additional and Advanced Exercises 117 16. Show lim g(x) œ lim xÄ$
xÄ$
x# 2x 3 2x 6
œ lim
xÄ$
(x 3)(x ") 2(x 3)
œ #, x Á 3. #
x 2x 3 2x 6 ,
Define the continuous extension of g(x) as G(x) œ œ
xÁ3 . We now prove the limit of g(x) as , xœ3
2
x Ä 3 exists and has the correct value. Step 1: ¹ x
#
2x 3 #x 6
2¹ % Ê %
(x 3)(x ") 2(x 3)
# % Ê %
x" #
# % , x Á $ Ê $ #% x $ #% .
Step 2: kx 3k $ Ê $ x 3 $ Ê $ $ x $ $. Then, $ $ œ $ #% Ê $ œ #%, or $ $ œ $ #% Ê $ œ #%. Choose $ œ #%. Then ! kx 3k $ Ê ¹x
#
2x 3 2x 6
2¹ % Ê lim
xÄ$
(x 3)(x ") #(x 3)
œ 2. Since the conditions of the continuity test hold for G(x),
g(x) can be continuously extended to G(x) at B œ 3. 17. (a) Let % ! be given. If x is rational, then f(x) œ x Ê kf(x) 0k œ kx 0k % Í kx 0k %; i.e., choose $ œ %. Then kx 0k $ Ê kf(x) 0k % for x rational. If x is irrational, then f(x) œ 0 Ê kf(x) 0k % Í ! % which is true no matter how close irrational x is to 0, so again we can choose $ œ %. In either case, given % ! there is a $ œ % ! such that ! kx 0k $ Ê kf(x) 0k %. Therefore, f is continuous at x œ 0. (b) Choose x œ c !. Then within any interval (c $ ß c $ ) there are both rational and irrational numbers. If c is rational, pick % œ #c . No matter how small we choose $ ! there is an irrational number x in (c $ ß c $ ) Ê kf(x) f(c)k œ k0 ck œ c
c #
œ %. That is, f is not continuous at any rational c 0. On
the other hand, suppose c is irrational Ê f(c) œ 0. Again pick % œ #c . No matter how small we choose $ ! there is a rational number x in (c $ ß c $ ) with kx ck œ kxk
c #
œ% Í
œ % Ê f is not continuous at any irrational c 0.
If x œ c 0, repeat the argument picking % œ nonzero value x œ c. 18. (a) Let c œ
c #
kc k #
œ
c # .
x
c #
Then kf(x) f(c)k œ kx 0k
3c #.
Therefore f fails to be continuous at any
m n
be a rational number in [0ß 1] reduced to lowest terms Ê f(c) œ "n . Pick % œ
" #n
œ %. Therefore f is discontinuous at x œ c, a rational number.
" #n .
No matter how small $ ! is taken, there is an irrational number x in the interval (c $ ß c $ ) Ê kf(x) f(c)k œ ¸0 "n ¸ œ
" n
(b) Now suppose c is an irrational number Ê f(c) œ 0. Let % 0 be given. Notice that number reduced to lowest terms with denominator 2 and belonging to [0ß 1]; denominator 3 belonging to [0ß 1];
" 4
and
[0ß 1]; etc. In general, choose N so that
" N
3 4
with denominator 4 in [0ß 1];
" 3
and
" 2 3 5, 5, 5
2 3
and
" #
is the only rational
the only rationals with 4 5
with denominator 5 in
% Ê there exist only finitely many rationals in [!ß "] having
denominator Ÿ N, say r" , r# , á , rp . Let $ œ min ekc ri k : i œ 1ß á ß pf . Then the interval (c $ ß c $ ) contains no rational numbers with denominator Ÿ N. Thus, 0 kx ck $ Ê kf(x) f(c)k œ kf(x) 0k œ kf(x)k Ÿ N" % Ê f is continuous at x œ c irrational.
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118
Chapter 2 Limits and Continuity
(c) The graph looks like the markings on a typical ruler when the points (xß f(x)) on the graph of f(x) are connected to the x-axis with vertical lines.
19. Yes. Let R be the radius of the equator (earth) and suppose at a fixed instant of time we label noon as the zero point, 0, on the equator Ê 0 1R represents the midnight point (at the same exact time). Suppose x" is a point on the equator “just after" noon Ê x" 1R is simultaneously “just after" midnight. It seems reasonable that the temperature T at a point just after noon is hotter than it would be at the diametrically opposite point just after midnight: That is, T(x" ) T(x" 1R) 0. At exactly the same moment in time pick x# to be a point just before midnight Ê x# 1R is just before noon. Then T(x# ) T(x# 1R) 0. Assuming the temperature function T is continuous along the equator (which is reasonable), the Intermediate Value Theorem says there is a point c between 0 (noon) and 1R (simultaneously midnight) such that T(c) T(c 1R) œ 0; i.e., there is always a pair of antipodal points on the earth's equator where the temperatures are the same. #
#
# # " 20. xlim f(x)g(x) œ xlim af(x) g(x)b‹ Šxlim af(x) g(x)b‹ “ ’af(x) g(x)b af(x) g(x)b “ œ "% ’Šxlim Äc Äc % Äc Äc œ "% ˆ$# a"b# ‰ œ #.
21. (a) At x œ 0: lim r (a) œ lim aÄ!
œ lim
1 (" a)
aÄ!
a Ä ! a ˆ" È1 a‰
At x œ 1: (b) At x œ 0:
lim
a Ä "b
œ
r (a) œ
" È1 a a 1 " È1 0
œ lim c aÄ!
1 (" a) a ˆ" È1 a‰
œ
" #
aÄ!
1 (1 a)
lim
a Ä "b a ˆ1 È1 a‰
lim r (a) œ lim c aÄ!
a Ä !c
È1 a
œ lim Š " a
" È1 a a
œ lim c aÄ!
" È1 a
‹ Š " È1 a ‹ a
œ lim
a Ä 1 a ˆ" È1 a‰ È1 a
œ lim c Š " a aÄ!
a a ˆ 1 È 1 a ‰
œ lim c aÄ!
œ
" " È0
œ1
" È1 a
‹ Š " È1 a ‹
" œ _ (because the " È1 a " œ _ (because the " È1 a
denominator is always negative); lim b r (a) œ lim b aÄ! aÄ! is always positive). Therefore, lim r (a) does not exist. aÄ!
At x œ 1:
lim r (a) œ lim b a Ä "b a Ä "
1 È 1 a a
œ
lim
"
a Ä 1b " È1 a
œ1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
denominator
Chapter 2 Additional and Advanced Exercises 119 (c)
(d)
22. f(x) œ x 2 cos x Ê f(0) œ 0 2 cos 0 œ 2 0 and f(1) œ 1 2 cos (1) œ 1 # 0. Since f(x) is continuous on [1ß !], by the Intermediate Value Theorem, f(x) must take on every value between [1 #ß #]. Thus there is some number c in [1ß !] such that f(c) œ 0; i.e., c is a solution to x 2 cos x œ 0. 23. (a) The function f is bounded on D if f(x) M and f(x) Ÿ N for all x in D. This means M Ÿ f(x) Ÿ N for all x in D. Choose B to be max ekMk ß kNkf . Then kf(x)k Ÿ B. On the other hand, if kf(x)k Ÿ B, then B Ÿ f(x) Ÿ B Ê f(x) B and f(x) Ÿ B Ê f(x) is bounded on D with N œ B an upper bound and M œ B a lower bound. (b) Assume f(x) Ÿ N for all x and that L N. Let % œ L # N . Since x lim f(x) œ L there is a $ ! such that Äx !
0 kx x! k $ Ê kf(x) Lk % Í L % f(x) L % Í L Í
LN #
f(x)
3L N # .
But L N Ê
LN #
Ê L
f(x) L
ML #
Í
3L M #
f(x)
ML # . As in part (b), 0 kx L M M, a contradiction. #
24. (a) If a b, then a b 0 Ê ka bk œ a b Ê max (aß b) œ
ab #
ka b k #
If a Ÿ b, then a b Ÿ 0 Ê ka bk œ (a b) œ b a Ê max (aß b) œ œ
2b #
f(x) L
x! k $
ab ab 2a # # œ # œ a. ka b k ab œ a # b b # a # #
œ
œ b.
(b) Let min (aß b) œ
ab #
ka b k #
LN #
N Ê N f(x) contrary to the boundedness assumption
f(x) Ÿ N. This contradiction proves L Ÿ N. (c) Assume M Ÿ f(x) for all x and that L M. Let % œ ML #
LN #
.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
120
Chapter 2 Limits and Continuity
25. lim œ xÄ0
sina" cos xb x
œ lim
œ lim
xÄ0
†
sin x
xÄ0 x
26.
lim
sin x
x Ä 0b sin Èx
œ
sina" cos xb " cos x
sin x " cos x
sin x
lim x Ä 0b B
†
†
" cos x x
Èx sin Èx
†
†
œ lim
x Èx
œ lim
sinasin xb sin x
28. lim
sinax# xb x
œ lim
sinax# xb x# x
† ax "b œ lim
sinax# %b x Ä 2 x2
œ lim
sinax# %b # x Ä 2 x %
† ax 2b œ lim
xÄ0
29. lim
xÄ0
sinˆÈx $‰ x9 xÄ9
30. lim
sin x x
xÄ0
sinasin xb sin x
†
sina" cos xb " cos x
† lim
" cos# x
x Ä 0 xa" cos xb
" Èx $
† lim
sin x
xÄ0 x
œ " † lim
œ " † " œ ".
sinax# xb x# x
† lim ax "b œ " † " œ "
sinax# %b # x Ä 2 x %
† lim ax 2b œ " † % œ %
xÄ0
sinˆÈx $‰ x Ä 9 Èx $
œ lim
xÄ0
œ " † lim b sin"Èx † lim b Èx œ " † ! † ! œ !. x Ä 0 Š Èx ‹ x Ä 0
sinasin xb x
xÄ0
œ lim
œ " † ˆ #! ‰ œ !.
27. lim
xÄ0
" cos x " cos x
†
xÄ0
xÄ2
sinˆÈx $‰ x Ä 9 Èx $
œ lim
† lim
"
x Ä 9 Èx $
œ"†
" '
œ
" '
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
sin# x
x Ä 0 xa" cos xb
CHAPTER 3 DIFFERENTIATION 3.1 THE DERIVATIVE OF A FUNCTION 1. Step 1: f(x) œ 4 x# and f(x h) œ 4 (x h)# f(x h) f(x) h
Step 2:
œ
c4 (x h)# d a4 x# b h
œ
a4 x# 2xh h# b 4 x# h
œ
2xh h# h
œ
h(2x h) h
œ 2x h Step 3: f w (x) œ lim (2x h) œ 2x; f w ($) œ 6, f w (0) œ 0, f w (1) œ 2 hÄ!
c(x h 1)# 1d c(x 1)# 1d h hÄ! 2xh h# 2h lim œ lim (2x h 2) h hÄ! hÄ!
2. F(x) œ (x 1)# 1 and F(x h) œ (x h 1)# " Ê Fw (x) œ lim ax# 2xh h# 2x 2h 1 1b ax# 2x 1 1b h
œ lim
hÄ!
œ
œ 2(x 1); Fw (1) œ 4, Fw (0) œ 2, Fw (2) œ 2 3. Step 1: g(t) œ
" t#
and g(t h) œ "
"
# # g(t h) g(t) œ (t h)h t h 2t h) 2t h œ h( (t h)# t# h œ (t h)# t#
Step 2:
2t h
Step 3: gw (t) œ lim
1 z #z
4. k(z) œ
and k(z h) œ
Œ
œ
œ
# # h Ä ! (t h) t
" (t h)#
h
2t t# †t#
1 (z h) 2(z h)
œ
(" z)(z h) lim (1 z h)z #(z h)zh hÄ!
œ
" 2z#
œ
t# (t h)# (t h)# †t#
œ
2 t$
t# at# 2th h# b (t h)# †t# †h
œ
œ
2th h# (t h)# t# h
2 ; gw (1) œ 2, gw (2) œ "4 , gw ŠÈ3‹ œ 3È 3
Ê kw (z) œ lim
Š
" (z h) " z ‹ #(z h) #z h
hÄ!
# z h z# zh lim z z zh 2(z h)zh hÄ!
h
œ lim
h Ä ! 2(z h)zh
œ lim
"
h Ä ! #(z h)z
; kw (") œ "# , kw (1) œ "# , kw ŠÈ2‹ œ "4
5. Step 1: p()) œ È3) and p() h) œ È3() h) Step 2:
p() h) p()) h
œ
œ
È3() h) È3) h
3h h ŠÈ3) 3h È3)‹
Step 3: pw ()) œ lim
œ
ŠÈ3) 3h È3)‹
œ
3 È3) 3h È3)
3
œ
h Ä ! È3) 3h È3)
†
h
3 È 3) È 3)
œ
3 2È 3 )
ŠÈ3) 3h È3)‹ ŠÈ3) 3h È3)‹
; pw (1) œ
6. r(s) œ È2s 1 and r(s h) œ È2(s h) 1 Ê rw (s) œ lim
hÄ!
œ lim
hÄ!
œ lim
ŠÈ2s h 1 È2s 1‹ h
†
2h
h Ä ! h ŠÈ2s 2h 1 È2s 1‹
œ
" È2s 1
; rw (0) œ 1, rw (1) œ
ŠÈ2s 2h 1 È2s 1‹ ŠÈ2s 2h 1 È2s 1‹
œ lim
" È3
6x# h 6xh# 2h$ h hÄ!
3 #È2
È2s 2h 1 È2s 1 h
œ
2 È2s 1 È2s 1
œ
2 2È2s 1
" È2
dy dx
h a6x# 6xh 2h# b h hÄ!
œ lim
, pw (3) œ "# , pw ˆ 32 ‰ œ
h Ä ! h ŠÈ2s 2h 1 È2s 1‹
2
7. y œ f(x) œ 2x$ and f(x h) œ 2(x h)$ Ê œ lim
(3) 3h) 3) h ŠÈ3) 3h È3)‹
(2s 2h 1) (2s 1)
œ lim
h Ä ! È2s 2h 1 È2s 1
, rw ˆ #" ‰ œ
3 2È 3
œ
2(x h)$ 2x$ h hÄ!
œ lim
2 ax$ 3x# h 3xh# h$ b 2x$ h hÄ!
œ lim
œ lim a6x# 6xh 2h# b œ 6x# hÄ!
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
122
Chapter 3 Differentiation
8. r œ œ
s$ #
1 Ê
”
œ lim
dr ds
(s h)$
#
t 2t1
Š
œ lim
(t b h)(2t b 1) c t(2t b 2h b 1) ‹ (2t b 2h b 1)(2t b 1)
h
œ
2t# t 2ht h 2t# 2ht t (2t 2h 1)(2t 1)h hÄ! " " (2t 1)(2t 1) œ (2t 1)#
dv dt
œ lim
œ
"
th“
ˆt "t ‰
# # th lim ht h(th h)t hÄ!
" Èq 1
11. p œ f(q) œ
œ
œ lim
h
" " t
h
th
œ
t# 1 t#
" È(q h) 1
Ê
h Ä ! (2t 2h 1)(2t 1)
Š
h(t h)t t (t h) ‹ (t h)t
h
hÄ!
œ1
" t#
Š È(q "h) 1 ‹ Š Èq" 1 ‹
œ lim
dp dq
"
œ lim
œ lim
h
hÄ!
h
hÄ!
Èq 1 Èq h 1
œ lim
h Ä ! hÈ q h 1 È q 1
h
hÄ!
3 # # s
h
hÄ!
h Ä ! (2t 2h 1)(2t 1)h
and f(q h) œ
3sh h# b œ
t ‰ Š 2(t bt bh)hb 1 ‹ ˆ 2t b 1
œ lim
ds dt
œ lim
# 1 lim t (thth)t hÄ!
Èq b 1 c Èq b h b 1 Œ Èq b h b 1 Èq b 1
œ
(t h)(2t 1) t(2t 2h 1) (2t 2h 1)(2t 1)h
hÄ!
œ lim
h
hÄ!
Ê
œ lim
œ lim
’(t h)
œ
th 2(th)1
and r(t h) œ
hÄ!
c(s h)$ 2d cs$ 2d " h # hlim Ä! h c3s# 3sh h# d " " œ # lim a3s# # hlim h Ä! hÄ!
h
hÄ!
" s$ 3s# h 3sh# h$ 2 s$ 2 # hlim h Ä!
9. s œ r(t) œ
10.
$
1• ’ s# 1“
œ
ˆÈ q 1 È q h 1 ‰ ˆ È q 1 È q h 1 ‰ 1) (q h 1) † ˆÈq 1 Èq h 1‰ œ lim hÈq h 1(qÈq 1 ˆÈ q 1 È q h 1 ‰ h Ä ! h Èq h 1 Èq 1 hÄ! h " lim œ lim Èq h 1 Èq 1 ˆÈq 1 Èq h 1‰ h Ä ! h È q h 1 È q 1 ˆÈ q 1 È q h 1 ‰ hÄ! " " œ È q 1 È q 1 ˆÈ q 1 È q 1 ‰ 2(q 1) Èq 1
dz dw
œ lim
œ lim œ
12.
" Š È3(w h) 2 h
hÄ!
"
È3w 2 ‹
ŠÈ3w 2 È3w 3h 2‹
œ lim
hÈ3w 3h 2 È3w 2
hÄ!
È3w 2 È3w 3h 2
œ lim
h Ä ! hÈ3w 3h 2 È3w 2
†
ŠÈ3w2È3w3h2‹ ŠÈ3w 2 È3w 3h 2‹
œ lim
(3w 2) (3w 3h 2)
œ lim
3
h Ä ! hÈ3w 3h 2 È3w 2 ŠÈ3w 2 È3w 3h 2‹ h Ä ! È3w 3h 2 È3w 2 ŠÈ3w 2 È3w 3h 2‹
œ
9 x
and f(x h) œ (x h)
œ
x(x h)# 9x x# (x h) 9(x h) x(x h)h
œ
h(x# xh 9) x(x h)h
14. k(x) œ
" #x
œ lim
hÄ!
œ lim
(# x) (2 x h) h(2 x)(2 x h)
hÄ!
œ lim
hÄ!
œ
x# xh 9 x(x h)
œ
w
œ
9 (x h)
Ê
f(x h) f(x) h
œ
’(x h)
x$ 2x# h xh# 9x x$ x# h 9x 9h x(x h)h
; f (x) œ
and k(x h) œ
" kw (2) œ 16
ds dt
3 È3w 2 È3w 2 ŠÈ3w 2 È3w 2‹
3 2(3w 2) È3w 2
13. f(x) œ x
15.
œ
# lim x xh 9 h Ä ! x(x h)
œ
x# 9 x#
9 9 (x b h) “ ’x x “
h
œ
œ1
x# h xh# 9h x(x h)h 9 x#
; m œ f w (3) œ 0
Š # "x h k(x h) k(x) œ lim h h hÄ! hÄ! h " " lim œ lim (2 x)(# x h) œ (2 x)# ; h Ä ! h(2 x)(2 x h) hÄ! " 2 (x h)
c(t h)$ (t h)# d at$ t# b h
3t# h 3th# h$ 2th h# h
Ê kw (x) œ lim
œ lim
hÄ!
œ lim
hÄ!
" #x‹
at$ 3t# h 3th# h$ b at# 2th h# b t$ t# h
h a3t# 3th h# 2t hb h
œ lim a3t# 3th h# 2t hb hÄ!
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 3.1 The Derivative of a Function œ 3t# 2t; m œ 16.
dy dx
ds ¸ dt tœ"
œ5
(x h 1)$ (x 1)$ h
œ lim
hÄ!
(x 1)$ 3(x ")# h 3(x 1)h# h$ (x 1)$ h
œ lim
hÄ!
œ lim c3(x 1)# 3(x 1)h h# d œ 3(x 1)# ; m œ hÄ!
17. f(x) œ œ
8 Èx 2
and f(x h) œ
8 ŠÈx 2 Èx h 2‹ hÈ x h 2 È x 2
†
8 È(x h) 2
f(x h) f(x) h
Ê
ŠÈx 2 Èx h 2‹
œ
ŠÈx 2 Èx h 2‹
œ
8h hÈx h 2 Èx 2 ŠÈx 2 Èx h 2‹
œ
8 Èx 2 Èx 2 ŠÈx 2 Èx 2‹
œ
œ3
dy dx ¹ x=#
È(x b h) c 2 Èx c 2 8
œ
8
h
8[(x 2) (x h 2)] hÈx h 2 Èx 2 ŠÈx 2 Èx h 2‹ 8
Ê f w (x) œ lim
h Ä ! Èx h 2 Èx 2 ŠÈx 2 Èx h 2‹
4 (x 2)Èx 2
; m œ f w (6) œ
4 4È 4
œ "# Ê the equation of the tangent
line at (6ß 4) is y 4 œ "# (x 6) Ê y œ "# x $ % Ê y œ "# x (. 18. gw (z) œ lim
ˆ1 È4 (z h)‰ Š1 È4 z‹ h
hÄ!
œ
h lim h Ä ! h ŠÈ4 z h È4 z‹
œ
†
h
hÄ!
(4 z h) (4 z) lim h Ä ! h ŠÈ4 z h È4 z‹ " œ "# 2È 4 3 "# z $# # Ê w
ŠÈ4 z h È4 z‹
œ lim
ŠÈ4 z h È4 z‹ ŠÈ4 z h È4 z‹ "
œ lim
h Ä ! ŠÈ4 z h È4 z‹
œ
" 2È 4 z
m œ gw (3) œ
Ê the equation of the tangent line at ($ß #) is w 2 œ "# (z 3)
Êwœ
œ "# z (# .
19. s œ f(t) œ 1 3t# and f(t h) œ 1 3(t h)# œ 1 3t# 6th 3h# Ê a1 3t# 6th 3h# b a1 3t# b h hÄ!
œ lim
20. y œ f(x) œ " œ lim
hÄ!
h h Ä ! x(x h)h
hÄ!
2È 4 ) 2È 4 ) h hÈ 4 ) È 4 ) h
œ
œ
" xh
Ê
" lim h Ä ! x(x h)
2 È4 () h)
4(% )) 4(% ) h)
œ lim
h Ä ! 2hÈ4 ) È4 ) h ŠÈ4 ) È4 ) h‹
œ
2 (4 )) Š2È4 )‹
œ
dy dx
œ
œ lim
" x#
" 3
Ê
" (4 ))È4 )
Ê
Šz h Èz h‹ ˆz Èz‰
hÄ!
h
œ 1 lim
(z h) z
h Ä ! h ŠÈz h Èz‹
hÄ!
Š1
dr ¸ d) )œ!
Ê
dr d)
È œ
È4 c ) c h È4 c ) 2
2
h
2
h Ä ! È4 ) È4 ) h ŠÈ4 ) È% ) h‹
œ
" 8
h Èz h Èz h hÄ!
œ lim
œ 1 lim
"
h
œ lim
œ lim
"
x h ‹ Š1 x ‹
hÄ!
dy dx ¹x= 3
f(t h) f(t) h
œ6
f(x h) f(x) h hÄ!
œ lim
22. w œ f(z) œ z Èz and f(z h) œ (z h) Èz h Ê œ lim
ds ¸ dt t="
œ lim
f() h) f()) œ lim h hÄ! hÄ! È È 2È4 ) #È% ) h Š2 % ) 2 4 ) h‹ lim † È Š2 4 ) #È4 ) h‹ h Ä ! hÈ 4 ) È 4 ) h
and f() h) œ
2 È4 )
hÄ!
and f(x h) œ 1
" " x xh œ lim h
21. r œ f()) œ œ lim
" x
œ lim (6t 3h) œ 6t Ê
ds dt
;
"
h Ä ! Èz h Èz
dw dz
œ lim
hÄ!
œ lim –1 hÄ!
œ"
" 2È z
Ê
f(z h) f(z) h Èz h Èz
dw ¸ dz zœ4
h
œ
†
ŠÈz h Èz‹ ŠÈz h Èz‹ —
5 4
" "
fazb faxb a x #b a z # b xz " z# x# 23. f w axb œ zlim œ zlim œ zlim œ zlim œ zlim Äx zx Ä x zx Ä x az xbaz #bax #b Ä x az xbaz #bax #b Ä x az #bax #b œ ax " #b#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
123
124
Chapter 3 Differentiation "
"
#
#
#
#
Òax "b az "bÓÒax "b az "bÓ ax "b az"b fazb faxb az"b ax"b 24. f w axb œ zlim œ zlim œ zlim œ zlim zx Ä x zx Äx Ä x az xbaz "b# ax "b# Äx az xbaz "b# ax "b# ax zbax z 2b "ax z 2b œ zlim œ zlim œ Ä x az xbaz "b# ax "b# Ä x a z " b # a x "b# z
"a#x #b a x "b %
œ
#ax "b a x "b %
œ
# a x "b $
x
gazb gaxb z a x "b x a z " b z x " zc" x" 25. gw axb œ zlim œ zlim œ zlim œ zlim œ zlim Äx zx Äx zx Ä x az xbaz "bax "b Ä x az xbaz "bax "b Ä x az "bax "b œ ax " "b# gazb gaxb 26. gw axb œ zlim œ zlim Äx zx Äx " " œ zlim œ #È x Ä x Èz Èx
ˆ" Èz‰ˆ" Èx‰ zx
œ zlim Äx
Èz Èx zx
†
Èz Èx Èz Èx
zx œ zlim Ä x az x bˆÈ z È x ‰
27. Note that as x increases, the slope of the tangent line to the curve is first negative, then zero (when x œ 0), then positive Ê the slope is always increasing which matches (b). 28. Note that the slope of the tangent line is never negative. For x negative, f#w (x) is positive but decreasing as x increases. When x œ 0, the slope of the tangent line to x is 0. For x 0, f#w (x) is positive and increasing. This graph matches (a). 29. f$ (x) is an oscillating function like the cosine. Everywhere that the graph of f$ has a horizontal tangent we expect f$w to be zero, and (d) matches this condition. 30. The graph matches with (c). 31. (a) f w is not defined at x œ 0, 1, 4. At these points, the left-hand and right-hand derivatives do not agree. For example, lim c xÄ!
f(x) f(0) x0
œ slope of line joining (%ß 0) and (!ß #) œ
" #
but lim b xÄ!
line joining (0ß 2) and ("ß 2) œ 4. Since these values are not equal, f w (0) œ lim
xÄ!
f(x) f(0) x0
f(x) f(0) x0
(b)
32. (a)
(b) Shift the graph in (a) down 3 units
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
œ slope of
does not exist.
Section 3.1 The Derivative of a Function 33.
(b) The fastest is between the 20th and 30th days; slowest is between the 40th and 50th days.
34. (a)
35. Left-hand derivative: For h 0, f(0 h) œ f(h) œ h# (using y œ x# curve) Ê œ lim c hÄ!
h# 0 h
œ lim c h œ 0; hÄ!
Right-hand derivative: For h 0, f(0 h) œ f(h) œ h (using y œ x curve) Ê œ lim b hÄ! Then lim c hÄ!
h0 h
lim
h Ä !c
œ lim b 1 œ 1; hÄ!
f(0 h) f(0) h
Á lim b hÄ!
f(0 h) f(0) h
lim
h Ä !b
œ lim c 0 œ 0; hÄ!
f(1 h) f(1) h
lim
h Ä !c
Right-hand derivative: When h !, 1 h 1 Ê f(1 h) œ 2(1 h) œ 2 2h Ê
Then lim c hÄ!
(2 2h)2 h
œ lim b hÄ!
f(1 h) f(1) h
2h h
È1 h " h
œ lim c hÄ!
lim
h Ä !b
ŠÈ1 h "‹ h
†
ŠÈ1 h "‹ ŠÈ1 h 1‹
œ lim c hÄ!
lim
h Ä !c
Then lim c hÄ!
(2h 1) " h f(1 h) f(1) h
38. Left-hand derivative:
lim
h Ä !c
Right-hand derivative: œ lim b hÄ! Then lim c hÄ!
h h(1 h)
œ lim b 2 œ 2; hÄ! Á lim b hÄ!
lim b
hÄ!
œ lim b hÄ!
f(1 h) f(1) h
f(1 h) f(1) h
f(1 h) f(") h
(1 h) " h ŠÈ1 h "‹
œ lim c hÄ!
Á lim b hÄ!
" È1 h 1
lim
h Ä !b
Ê the derivative f w (1) does not exist.
œ lim c
f(1 h) f(") h " 1h
f(1 h) f(1) h
f(" h) f(1) h
Right-hand derivative: When h 0, 1 h 1 Ê f(1 h) œ 2(1 h) 1 œ 2h 1 Ê œ lim b hÄ!
22 h
Ê the derivative f w (1) does not exist.
37. Left-hand derivative: When h 0, 1 h 1 Ê f(1 h) œ È1 h Ê œ lim c hÄ!
œ lim c hÄ!
œ lim b 2 œ 2; hÄ!
f(1 h) f(1) h
Á lim b hÄ!
f(0 h) f(0) h
Ê the derivative f w (0) does not exist.
36. Left-hand derivative: When h !, 1 h 1 Ê f(1 h) œ 2 Ê
œ lim b hÄ!
f(0 h) f(0) h
hÄ!
œ lim b hÄ!
(1 h) " h
œ lim c 1 œ 1;
" "‹ Š1 h h
hÄ!
œ lim b hÄ!
Š
1 (1 h) 1h ‹
h
œ 1; f(1 h) f(1) h
Ê the derivative f w (1) does not exist.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
œ #" ;
f("h)f(1) h
125
126
Chapter 3 Differentiation
39. (a) The function is differentiable on its domain $ Ÿ x Ÿ 2 (it is smooth) (b) none (c) none 40. (a) The function is differentiable on its domain # Ÿ x Ÿ 3 (it is smooth) (b) none (c) none 41. (a) The function is differentiable on $ Ÿ x 0 and ! x Ÿ 3 (b) none (c) The function is neither continuous nor differentiable at x œ 0 since lim c f(x) Á lim b f(x) xÄ! xÄ! 42. (a) f is differentiable on # Ÿ x 1, " x 0, 0 x 2, and 2 x Ÿ 3 (b) f is continuous but not differentiable at x œ 1: lim f(x) œ 0 exists but there is a corner at x œ 1 since x Ä 1
œ 3 and lim b f(" h)h f(1) œ 3 Ê f w (1) does not exist hÄ! hÄ! (c) f is neither continuous nor differentiable at x œ 0 and x œ 2: at x œ 0, lim c f(x) œ 3 but lim b f(x) œ 0 Ê lim f(x) does not exist; lim c
f(1 h) f(") h
xÄ!
xÄ0
xÄ!
at x œ 2, lim f(x) exists but lim f(x) Á f(2) xÄ#
xÄ#
43. (a) f is differentiable on " Ÿ x 0 and 0 x Ÿ 2 (b) f is continuous but not differentiable at x œ 0: lim f(x) œ 0 exists but there is a cusp at x œ 0, so f(0 h) f(0) h hÄ!
f w (0) œ lim
xÄ!
does not exist
(c) none 44. (a) f is differentiable on $ Ÿ x 2, 2 x 2, and 2 x Ÿ 3 (b) f is continuous but not differentiable at x œ 2 and x œ 2: there are corners at those points (c) none 45. (a) f w (x) œ lim
hÄ!
f(x h) f(x) h
œ lim
hÄ!
(x h)# ax# b h
œ lim
hÄ!
x# 2xh h# x# h
œ lim (2x h) œ 2x hÄ!
(b)
(c) yw œ 2x is positive for x 0, yw is zero when x œ 0, yw is negative when x 0 (d) y œ x# is increasing for _ x 0 and decreasing for ! x _; the function is increasing on intervals where yw 0 and decreasing on intervals where yw 0 f(x h) f(x) h hÄ!
46. (a) f w (x) œ lim
œ lim
hÄ!
Š xc" h h
1 ‹ x
œ lim
hÄ!
x (x h) x(x h)h
œ lim
"
h Ä ! x(x h)
œ
" x#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 3.1 The Derivative of a Function (b)
(c) yw is positive for all x Á 0, yw is never 0, yw is never negative (d) y œ "x is increasing for _ x 0 and ! x _ w
47. (a) Using the alternate formula for calculating derivatives: f (x) œ œ
$ $ lim z x z Ä x 3(z x)
œ
az# zx x# b lim (z x)3(z x) zÄx
œ
# # lim z zx3 x zÄx
f(x) lim f(z)z x zÄx # w
$
Š z3
œ zlim Äx
x$ 3 ‹
zx
œ x Ê f (x) œ x#
(b)
(c) yw is positive for all x Á 0, and yw œ 0 when x œ 0; yw is never negative (d) y œ
x$ 3
is increasing for all x Á 0 (the graph is horizontal at x œ 0) because y is increasing where yw 0; y is
never decreasing %
48. (a) Using the alternate œ zlim Äx
z% x% 4(z x)
œ
%
z x Œ4 4 f(z) f(x) form for calculating derivatives: f (x) œ zlim œ lim zx zx Äx zÄx (z x) az$ xz# x# z x$ b z$ xz# x# z x$ $ w lim œ zlim œ x Ê f (x) œ x$ 4(z x) 4 zÄx Äx
w
(b)
(c) yw is positive for x 0, yw is zero for x œ 0, yw is negative for x 0 (d) y œ
x% 4
is increasing on 0 x _ and decreasing on _ x 0 #
#
(xc) ax xc c b f(x) f(c) x c 49. yw œ xlim œ xlim œ xlim œ xlim ax# xc c# b œ 3c# . xc xc Äc Ä c xc Äc Äc The slope of the curve y œ x$ at x œ c is yw œ 3c# . Notice that 3c# 0 for all c Ê y œ x$ never has a negative slope. $
$
50. Horizontal tangents occur where yw œ 0. Thus, yw œ lim
hÄ!
œ lim
hÄ!
2 ŠÈx h Èx‹ h
†
ŠÈx h Èx‹ ŠÈx h Èx‹
œ lim
2È x h 2È x h
2((x h) x))
h Ä ! h ŠÈx h Èx‹
œ lim
2
h Ä ! Èx h Èx
œ
" Èx
.
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128
Chapter 3 Differentiation
Then yw œ 0 when 51. yw œ lim
hÄ!
œ lim
hÄ!
" Èx
œ 0 which is never true Ê the curve has no horizontal tangents.
a2(x h)# 13(x h) 5b a2x# 13x 5b h
4xh 2h# 13h h
œ lim
hÄ!
2x# 4xh 2h# 13x 13h 5 2x# 13x 5 h
œ lim (4x 2h 13) œ 4x 13, slope at x. The slope is 1 when 4x 13 œ " hÄ!
Ê 4x œ 12 Ê x œ 3 Ê y œ 2 † 3# 13 † 3 5 œ 16. Thus the tangent line is y 16 œ (1)(x 3) Ê y œ x "$ and the point of tangency is (3ß 16). 52. For the curve y œ Èx, we have yw œ lim
ŠÈx h Èx‹
hÄ!
"
œ lim
h Ä ! Èx h Èx
œ
" #Èx
h
†
ŠÈx h Èx‹ ŠÈx h Èx‹
h Ä ! ŠÈx h Èx‹ h
. Suppose ˆ+ß Èa‰ is the point of tangency of such a line and ("ß !) is the point
on the line where it crosses the x-axis. Then the slope of the line is " ; 2È a
(x h) x
œ lim
using the derivative formula at x œ a Ê
exist: its point of tangency is ("ß "), its slope is
Èa a1
œ
" #È a
œ
Èa 0 a (1)
œ
Èa a1
which must also equal
" Ê 2a œ a 1 Ê a œ 1. #Èa " # ; and an equation of the line is
Thus such a line does y1œ
" #
(x 1)
Ê y œ "# x "# . 53. No. Derivatives of functions have the intermediate value property. The function f(x) œ ÚxÛ satisfies f(0) œ 0 and f(1) œ 1 but does not take on the value "# anywhere in [!ß "] Ê f does not have the intermediate value
property. Thus f cannot be the derivative of any function on [!ß "] Ê f cannot be the derivative of any function on (_ß _).
54. The graphs are the same. So we know that for f(x) œ kxk , we have f w (x) œ
kx k x
.
55. Yes; the derivative of f is f w so that f w (x! ) exists Ê f w (x! ) exists as well. 56. Yes; the derivative of 3g is 3gw so that gw (7) exists Ê 3gw (7) exists as well. 57. Yes, lim
g(t)
t Ä ! h(t)
can exist but it need not equal zero. For example, let g(t) œ mt and h(t) œ t. Then g(0) œ h(0)
œ 0, but lim
g(t)
t Ä ! h(t)
œ lim
tÄ!
mt t
œ lim m œ m, which need not be zero. tÄ!
58. (a) Suppose kf(x)k Ÿ x# for " Ÿ x Ÿ 1. Then kf(0)k Ÿ 0# Ê f(0) œ 0. Then f w (0) œ lim œ lim
hÄ!
f(h) 0 h
œ lim
hÄ!
f(h) h .
For khk Ÿ 1, h# Ÿ f(h) Ÿ h# Ê h Ÿ
hÄ!
f(h) h
f(0 h) f(0) h
Ÿ h Ê f w (0) œ lim
hÄ!
f(h) h
œ0
by the Sandwich Theorem for limits. (b) Note that for x Á 0, kf(x)k œ ¸x# sin "x ¸ œ kx# k ksin xk Ÿ kx# k † 1 œ x# (since " Ÿ sin x Ÿ 1). By part (a), f is differentiable at x œ 0 and f w (0) œ 0.
59. The graphs are shown below for h œ 1, 0.5, 0.1. The function y œ y œ Èx so that
" #È x
œ lim
hÄ!
Èx h Èx h
" 2È x
. The graphs reveal that y œ
is the derivative of the function
Èx h Èx h
gets closer to y œ
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
" #È x
Section 3.1 The Derivative of a Function as h gets smaller and smaller.
60. The graphs are shown below for h œ 2, 1, 0.5. The function y œ 3x# is the derivative of the function y œ x$ so that 3x# œ lim
hÄ!
(xh)$ x$ h
. The graphs reveal that y œ
(xh)$ x$ h
gets closer to y œ 3x# as h
gets smaller and smaller.
61. Weierstrass's nowhere differentiable continuous function.
62-67. Example CAS commands: Maple: f := x -> x^3 + x^2 - x;
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129
130
Chapter 3 Differentiation
x0 := 1; plot( f(x), x=x0-5..x0+2, color=black, title="Section 3_1, #62(a)" ); q := unapply( (f(x+h)-f(x))/h, (x,h) ); # (b) L := limit( q(x,h), h=0 ); # (c) m := eval( L, x=x0 ); tan_line := f(x0) + m*(x-x0); plot( [f(x),tan_line], x=x0-2..x0+3, color=black, linestyle=[1,7], title="Section 3.1 #62(d)", legend=["y=f(x)","Tangent line at x=1"] ); Xvals := sort( [ x0+2^(-k) $ k=0..5, x0-2^(-k) $ k=0..5 ] ): # (e) Yvals := map( f, Xvals ): evalf[4](< convert(Xvals,Matrix) , convert(Yvals,Matrix) >); plot( L, x=x0-5..x0+3, color=black, title="Section 3.1 #62(f)" ); Mathematica: (functions and x0 may vary) (see section 2.5 re. RealOnly ): 0.78540 - 0.63662*cos(2*t) - 0.07074*cos(6*t) - 0.02546*cos(10*t) - 0.01299*cos(14*t); g := t -> piecewise( t cos(x)^2 + sin(x); ic := [x=Pi,y=1]; F := unapply( int( f(x), x ) + C, x ); eq := eval( y=F(x), ic ); solnC := solve( eq, {C} ); Y := unapply( eval( F(x), solnC ), x ); DEplot( diff(y(x),x) = f(x), y(x), x=0..2*Pi, [[y(Pi)=1]], color=black, linecolor=black, stepsize=0.05, title="Section 4.8 #103" ); Mathematica: (functions and values may vary) The following commands use the definite integral and the Fundamental Theorem of calculus to construct the solution of the initial value problems for exercises 103 - 105. Clear[x, y, yprime] yprime[x_] = Cos[x]2 Sin[x]; initxvalue = 1; inityvalue = 1; y[x_] = Integrate[yprime[t], {t, initxvalue, x}] inityvalue If the solution satisfies the differential equation and initial condition, the following yield True yprime[x]==D[y[x], x] //Simplify y[initxvalue]==inityvalue Since exercise 106 is a second order differential equation, two integrations will be required. Clear[x, y, yprime] y2prime[x_] = 3 Exp[x/2] 1; initxval = 0; inityval = 4; inityprimeval = 1; yprime[x_] = Integrate[y2prime[t],{t, initxval, x}] inityprimeval y[x_] = Integrate[yprime[t], {t, initxval, x}] inityval Verify that y[x] solves the differential equation and initial condition and plot the solution (red) and its derivative (blue). y2prime[x]==D[y[x], {x, 2}]//Simplify y[initxval]==inityval yprime[initxval]==inityprimeval Plot[{y[x], yprime[x]}, {x, initxval 3, initxval 3}, PlotStyle Ä {RGBColor[1,0,0], RGBColor[0,0,1]}] CHAPTER 4 PRACTICE EXERCISES 1. No, since f(x) œ x$ 2x tan x Ê f w (x) œ 3x# 2 sec# x 0 Ê f(x) is always increasing on its domain cos x 2. No, since g(x) œ csc x 2 cot x Ê gw (x) œ csc x cot x 2 csc# x œ sin #x
2 sin# x
œ sin"# x (cos x 2) 0
Ê g(x) is always decreasing on its domain 3. No absolute minimum because x lim (7 x)(11 3x)"Î$ œ _. Next f w (x) œ Ä_ (11 3x)"Î$ (7 x)(11 3x)#Î$ œ w
(11 3x) (7 x) (11 3x)#Î$
œ
4(1 x) (11 3x)#Î$
Ê x œ 1 and x œ
11 3
are critical points.
w
Since f 0 if x 1 and f 0 if x 1, f(1) œ 16 is the absolute maximum. 4. f(x) œ
ax b x# 1
Ê f w (x) œ
We require also that f(3) w
#a$x "bax $b ax # 1 b #
# a ax# 1b 2x(ax b) ab œ aaxax#2bx 1 b# ax # 1 b# œ 1. Thus " œ 3a8b Ê 3a b œ w
" ; f w (3) œ 0 Ê '% (*a 'b a) œ ! Ê &a $b œ !.
). Solving both equations yields a œ 6 and b œ 10. Now,
so that f œ ± ± ± ± . Thus f w changes sign at x œ $ from 1 1 3 1/3 positive to negative so there is a local maximum at x œ $ which has a value f(3) œ 1. f (x) œ
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Chapter 4 Practice Exercises
275
5. Yes, because at each point of [!ß "Ñ except x œ 0, the function's value is a local minimum value as well as a local maximum value. At x œ 0 the function's value, 0, is not a local minimum value because each open interval around x œ 0 on the x-axis contains points to the left of 0 where f equals 1. 6. (a) The first derivative of the function f(x) œ x$ is zero at x œ 0 even though f has no local extreme value at x œ 0. (b) Theorem 2 says only that if f is differentiable and f has a local extreme at x œ c then f w (c) œ 0. It does not assert the (false) reverse implication f w (c) œ 0 Ê f has a local extreme at x œ c. 7. No, because the interval 0 x 1 fails to be closed. The Extreme Value Theorem says that if the function is continuous throughout a finite closed interval a Ÿ x Ÿ b then the existence of absolute extrema is guaranteed on that interval. 8. The absolute maximum is k1k œ 1 and the absolute minimum is k0k œ 0. This is not inconsistent with the Extreme Value Theorem for continuous functions, which says a continuous function on a closed interval attains its extreme values on that interval. The theorem says nothing about the behavior of a continuous function on an interval which is half open and half closed, such as Ò"ß "Ñ, so there is nothing to contradict. 9. (a) There appear to be local minima at x œ 1.75 and 1.8. Points of inflection are indicated at approximately x œ 0 and x œ „ 1.
(b) f w (x) œ x( 3x& 5x% 15x# œ x# ax# 3b ax$ 5b. The pattern yw œ ± ± ± ± $È ! È$ È $ & $È indicates a local maximum at x œ 5 and local minima at x œ „ È3 . (c)
10. (a) The graph does not indicate any local extremum. Points of inflection are indicated at approximately x œ $% and x œ ".
(b) f w (x) œ x( 2x% 5
10 x$
œ x$ ax$ 2b ax( 5b . The pattern f w œ )( ± ± indicates (È $È ! & #
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276
Chapter 4 Applications of Derivatives a local maximum at x œ (È5 and a local minimum at x œ $È2 .
(c)
11. (a) g(t) œ sin# t 3t Ê gw (t) œ 2 sin t cos t 3 œ sin (2t) 3 Ê gw 0 Ê g(t) is always falling and hence must decrease on every interval in its domain. (b) One, since sin# t 3t 5 œ 0 and sin# t 3t œ 5 have the same solutions: f(t) œ sin# t 3t 5 has the same derivative as g(t) in part (a) and is always decreasing with f(3) 0 and f(0) 0. The Intermediate Value Theorem guarantees the continuous function f has a root in [$ß 0]. 12. (a) y œ tan ) Ê
dy d)
œ sec# ) 0 Ê y œ tan ) is always rising on its domain Ê y œ tan ) increases on every
interval in its domain (b) The interval 14 ß 1‘ is not in the tangent's domain because tan ) is undefined at ) œ
1 #
. Thus the tangent
need not increase on this interval. 13. (a) f(x) œ x% 2x# 2 Ê f w (x) œ 4x$ 4x. Since f(0) œ 2 0, f(1) œ 1 0 and f w (x) 0 for 0 Ÿ x Ÿ 1, we may conclude from the Intermediate Value Theorem that f(x) has exactly one solution when 0 Ÿ x Ÿ 1. È (b) x# œ 2 „ 4 8 0 Ê x# œ È3 1 and x 0 Ê x ¸ È.7320508076 ¸ .8555996772 #
14. (a) y œ
x x1
Ê yw œ
" (x 1)#
0, for all x in the domain of
x x1
Ê yœ
x x1
is increasing in every interval in
its domain (b) y œ x$ 2x Ê yw œ 3x# 2 0 for all x Ê the graph of y œ x$ 2x is always increasing and can never have a local maximum or minimum 15. Let V(t) represent the volume of the water in the reservoir at time t, in minutes, let V(0) œ a! be the initial amount and V(1440) œ a! (1400)(43,560)(7.48) gallons be the amount of water contained in the reservoir after the rain, where 24 hr œ 1440 min. Assume that V(t) is continuous on [!ß 1440] and differentiable on (!ß 1440). The Mean Value Theorem says that for some t! in (!ß 1440) we have Vw (t! ) œ œ
a! (1400)(43,560)(7.48) a! 1440
œ
456,160,320 gal 1440 min
V(1440) V(0) 1440 0
œ 316,778 gal/min. Therefore at t! the reservoir's volume
was increasing at a rate in excess of 225,000 gal/min. 16. Yes, all differentiable functions g(x) having 3 as a derivative differ by only a constant. Consequently, the d difference 3x g(x) is a constant K because gw (x) œ 3 œ dx (3x). Thus g(x) œ 3x K, the same form as F(x). x 1 x 1 x 1 œ 1 x 1 Ê x 1 differs from x 1 (x 1) x(1) d ˆ x ‰ d ˆ " ‰ œ (x " 1)# œ dx dx x 1 œ (x 1)# x1 .
17. No,
18. f w (x) œ gw (x) œ
2x ax # 1 b #
by the constant 1. Both functions have the same derivative
Ê f(x) g(x) œ C for some constant C Ê the graphs differ by a vertical shift.
19. The global minimum value of
" #
occurs at x œ #.
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Chapter 4 Practice Exercises
277
20. (a) The function is increasing on the intervals Ò$ß #Ó and Ò"ß #Ó. (b) The function is decreasing on the intervals Ò#ß !Ñ and Ð!ß "Ó. (c) The local maximum values occur only at x œ #, and at x œ #; local minimum values occur at x œ $ and at x œ " provided f is continuous at x œ !. 21. (a) t œ 0, 6, 12
(b) t œ 3, 9
(c) 6 t 12
(d) 0 t 6, 12 t 14
22. (a) t œ 4
(b) at no time
(c) 0 t 4
(d) 4 t 8
23.
24.
25.
26.
27.
28.
29.
30.
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278
Chapter 4 Applications of Derivatives
31.
32.
33. (a) yw œ 16 x# Ê yw œ ± ± Ê the curve is rising on (%ß %), falling on (_ß 4) and (%ß _) % % Ê a local maximum at x œ 4 and a local minimum at x œ 4; yww œ 2x Ê yww œ ± Ê the curve ! is concave up on (_ß !), concave down on (!ß _) Ê a point of inflection at x œ 0 (b)
34. (a) yw œ x# x 6 œ (x $)(x 2) Ê yw œ ± ± Ê the curve is rising on (_ß 2) and ($ß _), # $ falling on (#ß $) Ê local maximum at x œ 2 and a local minimum at x œ 3; yww œ 2x 1 Ê yww œ ± Ê concave up on ˆ "# ß _‰ , concave down on ˆ_ß "# ‰ Ê a point of inflection at x œ "# "Î# (b)
35. (a) yw œ 6x(x 1)(x 2) œ 6x$ 6x# 12x Ê yw œ ± ± ± Ê the graph is rising on ("ß !) " ! # and (#ß _), falling on (_ß 1) and (!ß #) Ê a local maximum at x œ 0, local minima at x œ 1 and x œ 2; yww œ 18x# 12x 12 œ 6 a3x# 2x 2b œ 6 Šx yww œ ± on
±
"È( $ È È Š 1 3 7 ß 1 3 7 ‹
"È( $
1 È7 3 ‹ Šx
Ê the curve is concave up on Š_ß
Ê points of inflection at x œ
1 È7 3 ‹
1 È7 3 ‹
Ê È7
and Š 1 3
1 „ È7 3
(b)
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ß _‹ , concave down
Chapter 4 Practice Exercises
279
36. (a) yw œ x# (6 4x) œ 6x# 4x$ Ê yw œ ± ± Ê the curve is rising on ˆ_ß #3 ‰, falling on ˆ #3 ß _‰ ! $Î# 3 ww Ê a local maximum at x œ # ; y œ 12x 12x# œ 12x(" x) Ê yww œ ± ± Ê concave up on ! " (!ß "), concave down on (_ß !) and ("ß _) Ê points of inflection at x œ 0 and x œ 1 (b)
37. (a) yw œ x% 2x# œ x# ax# 2b Ê yw œ ± ± ± Ê the curve is rising on Š_ß È2‹ and ! È# È # ŠÈ2ß _‹ , falling on ŠÈ2ß È2‹ Ê a local maximum at x œ È2 and a local minimum at x œ È2 ; yww œ 4x$ 4x œ 4x(x 1)(x 1) Ê yww œ ± ± ± Ê concave up on ("ß 0) and ("ß _), " ! " concave down on (_ß 1) and (0ß 1) Ê points of inflection at x œ 0 and x œ „ 1 (b)
38. (a) yw œ 4x# x% œ x# a4 x# b Ê yw œ ± ± ± Ê the curve is rising on (2ß 0) and (0ß 2), # ! # falling on (_ß 2) and (#ß _) Ê a local maximum at x œ 2, a local minimum at x œ 2; yww œ 8x 4x$ œ 4x a2 x# b Ê yww œ ± ± ± Ê concave up on Š_ß È2‹ and Š0ß È2‹ , concave ! È# È # down on ŠÈ2ß 0‹ and ŠÈ2ß _‹ Ê points of inflection at x œ 0 and x œ „ È2 (b)
39. The values of the first derivative indicate that the curve is rising on (!ß _) and falling on (_ß 0). The slope of the curve approaches _ as x Ä ! , and approaches _ as x Ä 0 and x Ä 1. The curve should therefore have a cusp and local minimum at x œ 0, and a vertical tangent at x œ 1.
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280
Chapter 4 Applications of Derivatives
40. The values of the first derivative indicate that the curve is rising on ˆ!ß "# ‰ and ("ß _), and falling on (_ß !) and ˆ "# ß "‰ . The derivative changes from positive to negative at x œ "# , indicating a local maximum there. The slope of the curve approaches _ as x Ä 0 and x Ä 1 , and approaches _ as x Ä 0 and as x Ä 1 , indicating cusps and local minima at both x œ 0 and x œ 1.
41. The values of the first derivative indicate that the curve is always rising. The slope of the curve approaches _ as x Ä 0 and as x Ä 1, indicating vertical tangents at both x œ 0 and x œ 1.
È33
42. The graph of the first derivative indicates that the curve is rising on Š!ß 17 16 on (_ß !) and xœ
17 È33 16
È È Š 17 16 33 ß 17 16 33 ‹
Ê a local maximum at x œ
17 È33 16
È33
‹ and Š 17 16
ß _‹ , falling
, a local minimum at
. The derivative approaches _ as x Ä 0 and x Ä 1, and approaches _ as x Ä 0 ,
indicating a cusp and local minimum at x œ 0 and a vertical tangent at x œ 1.
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Chapter 4 Practice Exercises
43. y œ
x1 x3
45. y œ
x# 1 x
œ1
œx
4 x3
" x
44. y œ
2x x5
œ2
46. y œ
x# x 1 x
10 x5
œx1
" x
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281
282
Chapter 4 Applications of Derivatives
47. y œ
x$ 2 #x
œ
49. y œ
x# 4 x# 3
œ1
51. lim
xÄ"
52. lim
54. lim
tan x x
œ
tan x
sin# x
sinamxb
x Ä 1Î#c
58.
xÄ"
axa" b" x Ä " bx
x Ä ! sinanxb
57. lim
œ lim
œ lim
# x Ä ! tanax b
56. lim
" x
" x# 3
xa "
x Ä ! x sin x
55. lim
x # $x % x"
b x Ä " x "
53. xlim Ä1
x# #
tan 1 1
#x $ "
œ
x% 1 x#
œ x#
50. y œ
x# x# 4
œ1
" x#
4 x# 4
œ&
a b
œ!
œ lim
sec# x
x Ä ! " cos x
œ lim
#sin x†cos x
# # x Ä ! #x sec ax b
œ lim
xÄ!
m cosamxb n cosanxb
œ
" ""
œ
" # sina#xb
œ lim
# # x Ä ! #x sec ax b
œ
cosa$xb
x Ä 1Î#c cosa(xb Èx cos x
59. lim acsc x cot xb œ lim
xÄ!
#cosa#xb
œ lim
# # # # # x Ä ! #x a#sec ax btanax b†#xb #sec ax b
œ
# ! #†"
œ"
m n
seca(xbcosa$xb œ lim
lim Èx sec x œ lim b x Ä !b xÄ! xÄ!
48. y œ
œ
! "
$sina$xb
œ lim
x Ä 1Î#c (sina(xb
œ
$ (
œ!
" cos x sin x
œ lim
sin x
x Ä ! cos x
œ
! "
œ!
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Chapter 4 Practice Exercises 60. lim ˆ x"% xÄ!
61.
#
œ lim Š " x%x ‹ œ lim a" x# b † xÄ!
xÄ!
" x%
œ lim a" x# b œ lim
ŠÈx# x " Èx# x‹ œ lim ŠÈx# x " Èx# x‹ † xÄ_ œ lim È # #x " È # xÄ_ x x" x x Notice that x œ Èx# for x ! so this is equivalent to lim
#x " x # x x " É x # x É # x x# $
œ lim xÄ_
$
lim Š x#x " x#x " ‹ œ lim xÄ_ xÄ_ "# " œ lim #% œ lim œ! x xÄ_ x Ä _ #x
"
% xÄ! x
xÄ!
xÄ_
œ lim xÄ_
62.
"‰ x#
œ"†_œ_
Èx# x "Èx# x È x# x " È x# x
# x" œ È"# È" œ " " É" x x"# É" x"
x $ a x # "b x $ a x # " b ax# "bax# "b
œ lim xÄ_
#x $ x% "
œ lim xÄ_
'x# %x $
œ lim xÄ_
"#x "#x#
63. (a) Maximize f(x) œ Èx È36 x œ x"Î# (36 x)"Î# where 0 Ÿ x Ÿ 36 Ê f w (x) œ
" #
x"Î# "# (36 x)"Î# (1) œ
È36 x Èx #Èx È36 x
Ê derivative fails to exist at 0 and 36; f(0) œ 6,
and f(36) œ 6 Ê the numbers are 0 and 36 (b) Maximize g(x) œ Èx È36 x œ x"Î# (36 x)"Î# where 0 Ÿ x Ÿ 36 Ê gw (x) œ
" #
x"Î# "# (36 x)"Î# (1) œ
È36 x Èx #Èx È36 x
Ê critical points at 0, 18 and 36; g(0) œ 6,
g(18) œ 2È18 œ 6È2 and g(36) œ 6 Ê the numbers are 18 and 18 64. (a) Maximize f(x) œ Èx (20 x) œ 20x"Î# x$Î# where 0 Ÿ x Ÿ 20 Ê f w (x) œ 10x"Î# 3# x"Î# œ
20 3x #È x
œ 0 Ê x œ 0 and x œ
œ
40È20 3È 3
Ê the numbers are
20 3
20 3
‰ É 20 ˆ are critical points; f(0) œ f(20) œ 0 and f ˆ 20 3 œ 3 20
and
40 3
.
(b) Maximize g(x) œ x È20 x œ x (20 x)"Î# where 0 Ÿ x Ÿ 20 Ê gw (x) œ Ê È20 x œ
" #
Ê xœ
the numbers must be 65. A(x) œ
" #
79 4
and
79 4 . " 4 .
The critical points are x œ
79 4
2È20 x 1 #È20 x
‰ and x œ 20. Since g ˆ 79 4 œ
(2x) a27 x# b for 0 Ÿ x Ÿ È27
Ê Aw (x) œ 3(3 x)(3 x) and Aw w (x) œ 6x. The critical points are 3 and 3, but 3 is not in the domain. Since Aw w (3) œ 18 0 and A ŠÈ27‹ œ 0, the maximum occurs at x œ 3 Ê the largest area is A(3) œ 54 sq units. 66. The volume is V œ x# h œ 32 Ê h œ 32 x# . The 32 ‰ # ˆ surface area is S(x) œ x 4x x# œ x# 128 x , where x 0 Ê Sw (x) œ
20 ‰ 3
2(x 4) ax# 4x 16b x#
Ê the critical points are 0 and 4, but 0 is not in the domain. Now Sw w (4) œ 2 256 4$ 0 Ê at x œ 4 there is a minimum. The dimensions 4 ft by 4 ft by 2 ft minimize the surface area.
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81 4
œ0 and g(20) œ 20,
283
284
Chapter 4 Applications of Derivatives #
67. From the diagram we have ˆ h# ‰ r# œ ŠÈ3‹ Ê r# œ
12h# 4
#
. The volume of the cylinder is #
V œ 1r# h œ 1 Š 12 4 h ‹ h œ
1 4
0 Ÿ h Ÿ 2È3 . Then Vw (h) œ
a12h h$ b , where 31 4
(2 h)(2 h)
Ê the critical points are 2 and 2, but 2 is not in the domain. At h œ 2 there is a maximum since Vw w (2) œ 31 0. The dimensions of the largest cylinder are radius œ È2 and height œ 2. 68. From the diagram we have x œ radius and y œ height œ 12 2x and V(x) œ "3 1x# (12 2x), where
0 Ÿ x Ÿ 6 Ê Vw (x) œ 21x(4 x) and Vw w (4) œ 81. The critical points are 0 and 4; V(0) œ V(6) œ 0 Ê x œ 4 gives the maximum. Thus the values of r œ 4 and h œ 4 yield the largest volume for the smaller cone.
‰ , where p is the profit on grade B tires and 0 Ÿ x Ÿ 4. Thus 69. The profit P œ 2px py œ 2px p ˆ 40510x x Pw (x) œ
2p (5 x)#
ax# 10x 20b Ê the critical points are Š5 È5‹, 5, and Š5 È5‹ , but only Š5 È5‹ is in
the domain. Now Pw (x) 0 for 0 x Š5 È5‹ and Pw (x) 0 for Š5 È5‹ x 4 Ê at x œ Š5 È5‹ there is a local maximum. Also P(0) œ 8p, P Š5 È5‹ œ 4p Š5 È5‹ ¸ 11p, and P(4) œ 8p Ê at x œ Š5 È5‹ there is an absolute maximum. The maximum occurs when x œ Š5 È5‹ and y œ 2 Š5 È5‹ , the units are hundreds of tires, i.e., x ¸ 276 tires and y ¸ 553 tires. 70. (a) The distance between the particles is lfatbl where fatb œ cos t cosˆt 1% ‰. Then, f w atb œ sin t sinˆt 1% ‰. Solving f w atb œ ! graphically, we obtain t ¸ "Þ"(), t ¸ %Þ$#!, and so on.
Alternatively, f w atb œ ! may be solved analytically as follows. f w atb œ sin’ˆt 1) ‰ 1) “ sin’ˆt 1) ‰ 1) “ œ ’sinˆt 1) ‰cos 1) cosˆt 1) ‰sin 1) “ ’sinˆt 1) ‰cos 1) cosˆt 1) ‰sin 1) “ œ #sin 1) cosˆt 1) ‰ so the critical points occur when cosˆt 1) ‰ œ !, or t œ
$1 )
k1. At each of these values, fatb œ „ cos $)1
¸ „ !Þ('& units, so the maximum distance between the particles is !Þ('& units.
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Chapter 4 Practice Exercises
285
(b) Solving cos t œ cos ˆt 1% ‰ graphically, we obtain t ¸ #Þ(%*, t ¸ &Þ)*!, and so on.
Alternatively, this problem can be solved analytically as follows. cos t œ cos ˆt 1% ‰ cos’ˆt 1) ‰ 1) “ œ cos’ˆt 1) ‰ 1) “ cosˆt 1) ‰cos 1) sinˆt 1) ‰sin 1) œ cosˆt 1) ‰cos 1) sinˆt 1) ‰sin 1) #sin ˆt 1) ‰sin 1) œ ! sin ˆt 1) ‰ œ ! tœ The particles collide when t œ
(1 )
(1 )
k1
¸ #Þ(%*. (plus multiples of 1 if they keep going.)
71. The dimensions will be x in. by "! #x in. by "' #x in., so Vaxb œ xa"! #xba"' #xb œ %x$ &#x# "'!x for ! x &. Then Vw axb œ "#x# "!%x "'! œ %ax #ba$x #!b , so the critical point in the correct domain is x œ #. This critical point corresponds to the maximum possible volume because Vw axb ! for ! x # and Vw axb ! for 2 x &. The box of largest volume has a height of 2 in. and a base measuring 6 in. by 12 in., and its volume is 144 in.$ Graphical support:
72. The length of the ladder is d" d# œ 8 sec ) 6 csc ). We wish to maximize I()) œ 8 sec ) 6 csc ) Ê Iw ()) œ 8 sec ) tan ) 6 csc ) cot ). Then Iw ()) œ 0 Ê 8 sin$ ) 6 cos$ ) œ 0 Ê tan ) œ
$ È 6
#
Ê
d" œ 4 É4 $È36 and d# œ $È36 É4 $È36 Ê the length of the ladder is about Š4 $È36‹ É4 $È36 œ Š4 $È36‹
$Î#
¸ "*Þ( ft.
73. g(x) œ 3x x$ 4 Ê g(2) œ 2 0 and g(3) œ 14 0 Ê g(x) œ 0 in the interval [#ß 3] by the Intermediate Value Theorem. Then gw (x) œ 3 3x# Ê xnb1 œ xn
3xn x$n 4 33xn#
; x! œ 2 Ê x" œ 2.22 Ê x# œ 2.196215, and
so forth to x& œ 2.195823345.
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286
Chapter 4 Applications of Derivatives
74. g(x) œ x% x$ 75 Ê g(3) œ 21 0 and g(4) œ 117 0 Ê g(x) œ 0 in the interval [$ß %] by the Intermediate Value Theorem. Then gw (x) œ 4x$ 3x# Ê xnb1 œ xn
x%n x$n 75 4xn$ 3xn#
; x! œ 3 Ê x" œ 3.259259
Ê x# œ 3.229050, and so forth to x& œ 3.22857729. 75.
' ax$ 5x 7b dx œ
76.
' Š8t$ t# t‹ dt œ 8t4% t6$ t## C œ 2t% t6$ t## C
77.
' ˆ3Èt t4# ‰ dt œ ' ˆ3t"Î# 4t# ‰ dt œ 3t$Î# 4t"1 C œ 2t$Î# 4t C
78.
' Š #È" t t3% ‹ dt œ ' ˆ #" t"Î# 3t% ‰ dt œ #" Œ t"Î# (3t$3) C œ Èt t"$ C
x% 4
5x# #
7x C
#
Š 3# ‹
" #
79. Let u œ r 5 Ê du œ dr
' ar dr5b
œ'
#
du u#
u" 1
œ ' u# du œ
C œ u" C œ ar " 5b C
80. Let u œ r È2 Ê du œ dr
'
6 dr
$
Šr È2‹
œ 6'
dr
$
Šr È2‹
œ 6'
du u$
81. Let u œ )# 1 Ê du œ 2) d) Ê
'
" #
3)È)# 1 d) œ ' Èu ˆ #3 du‰ œ
82. Let u œ 7 )2 Ê du œ 2) d) Ê
'È)
d) œ '
7 ) 2
" Èu
ˆ #" du‰ œ
" #
x$ a 1 x % b
"Î%
#
C
du œ ) d) 3 #
" #
3
ŠrÈ2‹
$Î# $Î# ' u"Î# du œ 3# Œ u$Î# C œ a ) # 1b C 3 C œ u #
du œ ) d)
"Î# ' u"Î# du œ #" Œ u"Î# C œ È7 )2 C " C œ u #
83. Let u œ 1 x% Ê du œ 4x$ dx Ê
'
#
œ 6' u$ du œ 6 Š u# ‹ C œ 3u# C œ
" 4
du œ x$ dx
dx œ ' u"Î% ˆ "4 du‰ œ
" 4
$Î% " $Î% ' u"Î% du œ 4" Œ u$Î% C œ 3" a1 x% b C 3 C œ 3 u 4
84. Let u œ 2 x Ê du œ dx Ê du œ dx
' (2 x)$Î& dx œ ' u$Î& ( du) œ ' u$Î& du œ u
)Î&
Š 85 ‹
85. Let u œ
'
s 10
sec# 10s
Ê du œ
" 10
C œ 58 u)Î& C œ 58 (2 x))Î& C
ds Ê 10 du œ ds
ds œ ' asec# ub (10 du) œ 10 ' sec# u du œ 10 tan u C œ 10 tan
86. Let u œ 1s Ê du œ 1 ds Ê
" 1
s 10
C
du œ ds
' csc# 1s ds œ ' acsc# ub ˆ 1" du‰ œ 1" ' csc# u du œ 1" cot u C œ 1" cot 1s C
87. Let u œ È2 ) Ê du œ È2 d) Ê
' csc È2) cot È2) d) œ '
" È2
du œ d)
(csc u cot u) Š È"2 du‹ œ
" È2
(csc u) C œ È"2 csc È2) C
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 4 Additional and Advanced Exercises ) 3
88. Let u œ
'
sec
) 3
tan
89. Let u œ
'
Ê du œ
x 4
) 3
" 3
d) Ê 3 du œ d)
d) œ ' (sec u tan u)(3 du) œ 3 sec u C œ 3 sec
Ê du œ
" 4
) 3
C
dx Ê 4 du œ dx
2u ‰ dx œ ' asin# ub (4 du) œ ' 4 ˆ 1 cos du œ 2' (1 cos 2u) du œ 2 ˆu # œ 2u sin 2u C œ 2 ˆ x4 ‰ sin 2 ˆ x4 ‰ C œ x# sin x# C
sin#
x 4
90. Let u œ
'
cos#
œ
x #
91. y œ '
x # x # " #
Ê du œ
" #
2u ‰ dx œ ' acos# ub (2 du) œ ' 2 ˆ 1 cos du œ ' (1 cos 2u) du œ u #
x# " x#
dx œ ' a1 x# b dx œ x x" C œ x
y œ 1 when x œ 1 Ê
" x
œ ' Š15Èt
3 Èt ‹
" 3
2
1 1
"‰ x#
Ê
" x
C; y œ 1 when x œ 1 Ê 1
dx œ ' ax# 2 x# b dx œ
C œ 1 Ê C œ 3" Ê y œ
$
x 3
x$ 3
2x x" C œ
2x
dt œ ' ˆ15t"Î# 3t"Î# ‰ dt œ 10t$Î# 6t"Î# C; dr dt œ &Î#
œ 4t&Î# 4t$Î# 8t C; r œ 0 when t œ 1 Ê 4(1) r œ 4t&Î# 4t$Î# 8t d# r dt#
C
dr dt
" x
C œ 1
x$ 3
2x
" x
C;
" 3
œ 8 when t œ 1
10t$Î# 6t"Î# 8 Ê r œ ' ˆ10t$Î# 6t"Î# 8‰ dt
4(1)$Î# 8(1) C" œ 0 Ê C" œ 0. Therefore,
œ ' cos t dt œ sin t C; rw w œ 0 when t œ 0 Ê sin 0 C œ 0 Ê C œ 0. Thus, dr dt
1 1
1
Ê 10(1)$Î# 6(1)"Î# C œ 8 Ê C œ 8. Thus
94.
sin 2u #
sin x C
# 92. y œ ' ˆx x" ‰ dx œ ' ˆx# 2
dr dt
C
dx Ê 2 du œ dx
Ê C œ 1 Ê y œ x
93.
sin 2u ‰ #
œ ' sin t dt œ cos t C" ; rw œ 0 when t œ 0 Ê 1 C" œ 0 Ê C" œ 1. Then
d# r dt# œ sin t dr dt œ cos t
1
Ê r œ ' (cos t 1) dt œ sin t t C# ; r œ 1 when t œ 0 Ê 0 0 C# œ 1 Ê C# œ 1. Therefore, r œ sin t t 1 CHAPTER 4 ADDITIONAL AND ADVANCED EXERCISES 1. If M and m are the maximum and minimum values, respectively, then m Ÿ f(x) Ÿ M for all x − I. If m œ M then f is constant on I. 3x 6, 2 Ÿ x 0 has an absolute minimum value of 0 at x œ 2 and an absolute 9 x# , 0 Ÿ x Ÿ 2 maximum value of 9 at x œ 0, but it is discontinuous at x œ 0.
2. No, the function f(x) œ œ
3. On an open interval the extreme values of a continuous function (if any) must occur at an interior critical point. On a half-open interval the extreme values of a continuous function may be at a critical point or at the closed endpoint. Extreme values occur only where f w œ 0, f w does not exist, or at the endpoints of the interval. Thus the extreme points will not be at the ends of an open interval. 4. The pattern f w œ ± ± ± ± indicates a local maximum at x œ 1 and a local " # $ % minimum at x œ 3.
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287
288
Chapter 4 Applications of Derivatives
5. (a) If yw œ 6(x 1)(x 2)# , then yw 0 for x 1 and yw 0 for x 1. The sign pattern is f w œ ± ± Ê f has a local minimum at x œ 1. Also yww œ 6(x 2)# 12(x 1)(x 2) " # œ 6(x 2)(3x) Ê yw w 0 for x 0 or x 2, while yww 0 for 0 x 2. Therefore f has points of inflection at x œ 0 and x œ 2. There is no local maximum. (b) If yw œ 6x(x 1)(x 2), then yw 0 for x 1 and 0 x 2; yw 0 for " x 0 and x 2. The sign sign pattern is yw œ ± ± ± . Therefore f has a local maximum at x œ 0 and " ! # È7
local minima at x œ 1 and x œ 2. Also, yww œ ") ’x Š 1 $ 1 È 7 $
x
1 È 7 $
È7
‹“ ’x Š 1 $
‹“ , so yww 0 for
and yww 0 for all other x Ê f has points of inflection at x œ
6. The Mean Value Theorem indicates that
f(6) f(0) 60
1 „È 7 $
.
œ f w (c) Ÿ 2 for some c in (0ß 6). Then f(6) f(0) Ÿ 12
indicates the most that f can increase is 12. 7. If f is continuous on [aß c) and f w (x) Ÿ 0 on [aß c), then by the Mean Value Theorem for all x − [aß c) we have f(c) f(x) cx
Ÿ 0 Ê f(c) f(x) Ÿ 0 Ê f(x) f(c). Also if f is continuous on (cß b] and f w (x) 0 on (cß b], then for f(x) f(c) xc
all x − (cß b] we have
0 Ê f(x) f(c) 0 Ê f(x) f(c). Therefore f(x) f(c) for all x − [aß b].
8. (a) For all x, (x 1)# Ÿ 0 Ÿ (x 1)# Ê a1 x# b Ÿ 2x Ÿ a1 x# b Ê "# Ÿ (b) There exists c − (aß b) such that Ê kf(b) f(a)k Ÿ
" #
c 1 c#
œ
f(b) f(a) ba
f(a) ¸ c ¸ Ê ¹ f(b)b a ¹ œ 1 c# Ÿ
" #
x 1 x#
Ÿ
" #
.
, from part (a)
kb ak .
9. No. Corollary 1 requires that f w (x) œ 0 for all x in some interval I, not f w (x) œ 0 at a single point in I. 10. (a) h(x) œ f(x)g(x) Ê hw (x) œ f w (x)g(x) f(x)gw (x) which changes signs at x œ a since f w (x), gw (x) 0 when x a, f w (x), gw (x) 0 when x a and f(x), g(x) 0 for all x. Therefore h(x) does have a local maximum at x œ a. (b) No, let f(x) œ g(x) œ x$ which have points of inflection at x œ 0, but h(x) œ x' has no point of inflection (it has a local minimum at x œ 0). 11. From (ii), f(1) œ lim
xÄ „_
f(x) œ
12.
dy dx
œ 0 Ê a œ 1; from (iii), either 1 œ x lim f(x) or 1 œ x Ä lim f(x). In either case, Ä_ _
x" # x Ä „ _ bx cx #
1 "x x c 2x xÄ „_
lim
" a bc#
œ ! and if c œ 0,
œ 3x# 2kx 3 œ 0 Ê x œ
1 "x bx c 2x xÄ „_ 1 x" then lim bx 2x xÄ „_
œ
lim
lim
2k „ È4k# 36 6
œ " Ê b œ 0 and c œ ". For if b œ ", then œ
lim
xÄ „_
1 x" 2 x
œ „ _. Thus a œ 1, b œ 0, and c œ 1.
Ê x has only one value when 4k# 36 œ 0 Ê k# œ 9 or
k œ „ 3. 13. The area of the ?ABC is A(x) œ w
where 0 Ÿ x Ÿ 1. Thus A (x) œ
" #
(2) È1 x# œ a1 x# b
x È 1 x#
"Î#
,
Ê 0 and „ 1 are
critical points. Also A a „ 1b œ 0 so A(0) œ 1 is the maximum. When x œ 0 the ?ABC is isosceles since AC œ BC œ È2 .
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 4 Additional and Advanced Exercises f (c h) f (c) œ f ww (c) Ê for % œ "# kf ww (c)k 0 h hÄ0 Ê ¹ f (ch)h f (c) f ww (c)¹ "# kf ww (c)k . Then f w (c) œ w
14. lim
w
w
w
3 #
f ww (c)
0 Ê "# kf ww (c)k
f (c h) f ww (c) "# kf ww (c)k . If f ww (c) 0, then h f (c h) "# f ww (c) 0; likewise if f ww (c) 0, then 0 "# h w
Ê f ww (c) "# kf ww (c)k Ê
there exists a $ 0 such that 0 khk $
w
w
f (c h) h
" #
f ww (c)
w
kf ww (c)k
kf ww (c)k œ f ww (c) f ww (c)
f (c h) h w
3 #
f ww (c).
(a) If f ww (c) 0, then $ h 0 Ê f (c h) 0 and 0 h $ Ê f w (c h) 0. Therefore, f(c) is a local maximum. (b) If f ww (c) 0, then $ h 0 Ê f w (c h) 0 and 0 h $ Ê f w (c h) 0. Therefore, f(c) is a local minimum. 15. The time it would take the water to hit the ground from height y is É 2y g , where g is the acceleration of gravity. The product of time and exit velocity (rate) yields the distance the water travels: È64(h y) œ 8 É 2 ahy y# b D(y) œ É 2y g g are critical points. Now D(0) œ 0,
D ˆ h# ‰
16. From the figure in the text, tan (" )) œ give
ba h
œ
tan " 1 ha tan " a h
œ
h tan " a h a tan "
œ
"Î#
, 0 Ÿ y Ÿ h Ê Dw (y) œ 4 É g2 ahy y# b
8h Èg
ba h ;
"Î#
(h 2y) Ê 0,
and D(h) œ 0 Ê the best place to drill the hole is at y œ
tan (" )) œ
tan " tan ) 1 tan " tan )
. Solving for tan " gives tan " œ
; and tan ) œ
bh h# a(b a)
a h
h #
h #
and h
.
. These equations
or
#
ah a(b a)b tan " œ bh. Differentiating both sides with respect to h gives 2h tan " ah# a(b a)b sec# "
d" dh
œ b. Then
d" dh
bh œ 0 Ê 2h tan " œ b Ê 2h Š h# a(b a) ‹ œ b
Ê 2bh# œ bh# ab(b a) Ê h# œ a(b a) Ê h œ Èa(a b) . 17. The surface area of the cylinder is S œ 21r# 21rh. From the diagram we have Rr œ H H h Ê h œ RH R rH and S(r) œ 21r(r h) œ 21r ˆr H r HR ‰ œ 21 ˆ1 HR ‰ r# 21Hr, where 0 Ÿ r Ÿ R.
Case 1: H R Ê S(r) is a quadratic equation containing the origin and concave upward Ê S(r) is maximum at r œ R. Case 2: H œ R Ê S(r) is a linear equation containing the origin with a positive slope Ê S(r) is maximum at r œ R. Case 3: H R Ê S(r) is a quadratic equation containing the origin and concave downward. Then dS H‰ dS H‰ RH ˆ ˆ dr œ 41 1 R r 21H and dr œ 0 Ê 41 1 R r 21H œ 0 Ê r œ 2(H R) . For simplification we let r‡ œ
RH 2(H R)
.
(a) If R H 2R, then 0 H 2R Ê H 2(H R) Ê (b) If H œ 2R, then r‡ œ
#
2R 2R
RH 2(H R)
R which is impossible.
œ R Ê S(r) is maximum at r œ R.
(c) If H 2R, then 2R H 2H Ê H 2(H R) Ê S(r) is a maximum at r œ r‡ œ
RH 2(H R)
H 2(H R)
1 Ê
RH 2(H R)
R Ê r‡ R. Therefore,
.
Conclusion: If H − (0ß R] or H œ 2R, then the maximum surface area is at r œ R. If H − (Rß 2R), then r R which is not possible. If H − (2Rß _), then the maximum is at r œ r‡ œ 2(HRH R) . 18. f(x) œ mx 1
" x
Ê f w (x) œ m
" x#
and f w w (x) œ
2 x$
0 when x 0. Then f w (x) œ 0 Ê x œ
minimum. If f Š È"m ‹ 0, then Èm 1 Èm œ 2Èm 1 0 Ê m
" 4
" Èm
yields a
. Thus the smallest acceptable value
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
289
290
Chapter 4 Applications of Derivatives
for m is 19. (a) (b) (c)
" 4
.
lim
xÄ!
#sina&xb $x
œ lim
xÄ!
x x Ä ! sin# È#x
lim
x Ä 1/2
"
xÄ!
(g) (h)
cos$ x #
sinax# b
lim x Ä ! xsin x lim
xÄ!
x$ )
20. (a) x lim Ä_ (b) x lim Ä_
"! $
$sina&xbsina$xb &cosa&xbcosa$xb $cosa$xb
xÄ!
È #x
" œ lim #sinÈ#x cosÈ#x xÄ! È#x
xÄ!
sinŠ#È#x‹
œ
& $ "
È#x
œ lim
x Ä ! cosŠ#È#x‹ È# #x
"sin x cos x
lim " cos#x x Ä ! " sec x
œ
cos x
œ lim
œ !Þ
x Ä 1/2 sin x
lim " cos# x x Ä ! tan x
œ
lim cos x# " x Ä ! tan x
œ lim
sin x
# x Ä ! # tan x sec x
œ lim
sin x
xÄ!
# sin x cos$ x
œ
œ #" #x cosax# b
œ lim
x Ä ! xcos xsin x
sec x " x#
lim # x Ä # x %
†"œ
" #
œ
x Ä 1/2
œ
"! $
œ
œ lim
œ lim
asec x tan xb œ lim
lim x sin x x Ä ! x tan x œ lim
(f)
sina&xbcosa$xb sina$xb
lim x csc# È#x œ lim x Ä ! cosŠ#È#x‹†#
(e)
xÄ! $
xÄ!
xÄ!
"! sina&xb a&xb
œ lim
lim sina&xbcota$xb œ lim
xÄ!
œ lim (d)
#sina&xb $ & a &x b
œ lim
xÄ!
œ lim
xÄ#
sec x tan x #x
xÄ!
ax #bax# #x %b ax #bax #b
œ x lim Ä_
#
sec x tan x sec x #
xÄ!
x # #x % x#
œ lim
xÄ#
œ x lim œ x lim Ä _ Èx & Ä_
#x x (È x
$
œ lim
Èx & Èx Èx
Èx & Èx &
a#x# bsinax# b #cosax# b xsin x#cos x
œ lim
É" x& " È&x
#x x x( x x
È œ x lim Ä_
œ
# " (É x"
œ œ
" "
œ"
œ
# "!
œ
"! #
# #
œ"
œ
%%% %
" #
œ$
œ#
21. (a) The profit function is Paxb œ ac exbx aa bxb œ ex# ac bbx a. Pw axb œ #ex c b œ ! Ê x œ c#eb . Pww axb œ #e ! if e ! so that the profit function is maximized at x œ c #e b . (b) The price therefore that corresponds to a production level yeilding a maximum profit is p¹
xœ c#eb
œ c eˆ c #e b ‰ œ
c b #
dollars. #
(c) The weekly profit at this production level is Paxb œ eˆ c #e b ‰ ac bbˆ c #e b ‰ a œ
ac b b # %e #
a.
(d) The tax increases cost to the new profit function is Faxb œ ac exbx aa bx txb œ ex ac b tbx a. bc cbt ww Now Fw axb œ #ex c b t œ ! when x œ t # e œ #e . Since F axb œ #e ! if e !, F is maximized when x œ c #be t units per week. Thus the price per unit is p œ c eˆ c #be t ‰ œ c #b t dollars. Thus, such a tax increases the cost per unit by
cbt #
The x-intercept occurs when
" x
cb #
œ
t #
dollars if units are priced to maximize profit.
22. (a)
$œ!Ê
(b) By Newton's method, xn" œ xn œ xn xn
$x#n
œ #xn
$xn#
faxn b f ax n b . w
" x
œ $ Ê x œ $" .
Here f w axn b œ x# n œ
" x#n .
" $
So xn" œ xn xn" œ xn Š x"n $‹x#n x# n
œ xn a# $xn b.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 4 Additional and Advanced Exercises 23. x" œ x! and
a q" x!
fax! b f w ax ! b
q
qx!q xq! a qxq!"
x a œ x! ! q" œ qx
!
with weights m! œ
In the case where x! œ
a xq!"
q" q
q
x aq "b a œ ! q" œ x! Š q q " ‹
a Š"‹ xq!" q
qx!
and m" œ "q .
we have xq! œ a and x" œ
q" a a " q" Š q ‹ q" Š q ‹ x! x!
œ
291
so that x" is a weighted average of x!
q" a q" Š q x!
q" ‹ œ
a q" . x! #
dy d y 24. We have that ax hb# ay hb# œ r# and so #ax hb #ay hb dy dx œ ! and # # dx #ay hb dx# œ ! hold. dy x y dx dy . " dx
dy Thus #x #y dy dx œ #h #h dx , by the former. Solving for h, we obtain h œ #
d y equation yields # # dy dx #y dx# #Œ
dy x y dx dy " dx
œ !. Dividing by 2 results in "
Substituting this into the second
dy dx
#
y ddxy# Œ
dy x y dx dy " dx
œ !.
25. (a) aatb œ sww atb œ k ak !b Ê sw atb œ kt C" , where sw a!b œ )) Ê C" œ )) Ê sw atb œ kt )). So satb œ kt# #
kt# #
))t C# where sa!b œ ! Ê C# œ ! so satb œ
))t œ "!!. Solving for t we obtain t œ
kŠ ))
È))# #!!k ‹ k ))# #!!
so that k œ
)) œ ! or kŠ ))
)) „ È))# #!!k . k
È))# #!!k ‹ k
kt# #
))t. Now satb œ "!! when
At such t we want sw atb œ !, thus
)) œ !. In either case we obtain ))# #!!k œ !
¸ $)Þ(# ft/sec# .
(b) The initial condition that sw a!b œ %% ft/sec implies that sw atb œ kt %% and satb œ w
The car is stopped at a time t such that s atb œ kt %% œ ! Ê t œ ‰ sˆ %% k
œ
k ˆ %% ‰# # k
‰ %%ˆ %% k
œ
%%# #k
œ
*') k
œ
‰ *')ˆ #!! ))#
%% k .
kt# #
%%t where k is as above.
At this time the car has traveled a distance
œ #& feet. Thus halving the initial velocity quarters
stopping distance. 26. haxb œ f # axb g# axb Ê hw axb œ #faxbf w axb #gaxbgw axb œ #faxbf w axb gaxbgw axb‘ œ #faxbgaxb gaxbafaxbb‘ œ # † ! œ !. Thus haxb œ c, a constant. Since ha!b œ &, haxb œ & for all x in the domain of h. Thus ha"!b œ &. 27. Yes. The curve y œ x satisfies all three conditions since
dy dx
œ " everywhere, when x œ !, y œ !, and
d# y dx#
œ ! everywhere.
28. yw œ $x# # for all x Ê y œ x$ #x C where " œ "$ # † " C Ê C œ % Ê y œ x$ #x %. 29. sww atb œ a œ t# Ê v œ sw atb œ maximum for this t‡ . Since satb t‡ œ a$Cb"Î$ . So ÊCœ
a%bb$Î% $ .
a$Cb"Î$ ‘% 12
t$ w ‡ ‡ $ C. We seek v! œ s a!b œ C. We know that sat b œ b for some t and s is at a % % œ 12t Ct k and sa!b œ ! we have that satb œ 12t Ct and also sw at‡ b œ ! so that
Ca$Cb"Î$ œ b Ê a$Cb"Î$ ˆC
Thus v! œ sw a!b œ
a%bb$Î% $
œ
#È# $Î% . $ b
$C ‰ "#
œ b Ê a$Cb"Î$ ˆ $%C ‰ œ b Ê $"Î$ C%Î$ œ
30. (a) sww atb œ t"Î# t"Î# Ê vatb œ sw atb œ #$ t$Î# #t"Î# k where va!b œ k œ (b) satb œ
% &Î# "& t
% %$ t$Î# %$ t k# where sa!b œ k# œ "& . Thus satb œ
% # $Î# #t"Î# $ Ê vatb œ $ t % &Î# % %$ t$Î# %$ t "& . "& t
%b $
%$ Þ
31. The graph of faxb œ ax# bx c with a ! is a parabola opening upwards. Thus faxb ! for all x if faxb œ ! for at most one real value of x. The solutions to faxb œ ! are, by the quadratic equation #
#b „ Éa#bb# %ac . #a
Thus we require
#
a#bb %ac Ÿ ! Ê b ac Ÿ !. 32. (a) Clearly faxb œ aa" x b" b# Þ Þ Þ aan x bn b# ! for all x. Expanding we see faxb œ aa#" x# #a" b" x b"# b Þ Þ Þ aan# x# #an bn x bn# b œ aa#" a## Þ Þ Þ an# bx# #aa" b" a# b# Þ Þ Þ an bn bx ab"# b## Þ Þ Þ bn# b !.
Thus aa" b" a# b# Þ Þ Þ an bn b# aa#" a## Þ Þ Þ an# bab"# b## Þ Þ Þ bn# b Ÿ ! by Exercise 31. Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
292
Chapter 4 Applications of Derivatives
Thus aa" b" a# b# Þ Þ Þ an bn b# Ÿ aa#" a## Þ Þ Þ an# bab"# b## Þ Þ Þ bn# b. (b) Referring to Exercise 31: It is clear that faxb œ ! for some real x Í b# %ac œ !, by quadratic formula. Now notice that this implies that faxb œ aa" x b" b# Þ Þ Þ aan x bn b# œ aa#" a## Þ Þ Þ an# bx# #aa" b" a# b# Þ Þ Þ an bn bx ab"# b## Þ Þ Þ bn# b œ ! Í aa" b" a# b# Þ Þ Þ an bn b# aa#" a## Þ Þ Þ an# bab"# b## Þ Þ Þ bn# b œ !
Í aa" b" a# b# Þ Þ Þ an bn b# œ aa#" a## Þ Þ Þ an# bab"# b## Þ Þ Þ bn# b But now faxb œ ! Í ai x bi œ ! for all i œ "ß #ß Þ Þ Þ ß n Í ai x œ bi œ ! for all i œ "ß #ß Þ Þ Þ ß n.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
CHAPTER 5 INTEGRATION 5.1 ESTIMATING WITH FINITE SUMS 1. faxb œ x#
Since f is increasing on Ò!ß "Ó, we use left endpoints to obtain lower sums and right endpoints to obtain upper sums.
"
#
iœ! $
" #
œ "# Š!# ˆ "# ‰ ‹ œ
#
" 4
œ 4" Š!# ˆ 4" ‰ ˆ #" ‰ ˆ 4$ ‰ ‹ œ
(a) ˜x œ
"! #
œ
" #
and xi œ i˜x œ
i #
Ê a lower sum is !ˆ #i ‰ †
(b) ˜x œ
"! %
œ
" %
and xi œ i˜x œ
i %
Ê a lower sum is !ˆ 4i ‰ †
(c) ˜x œ
"! #
œ
" #
and xi œ i˜x œ
i #
Ê an upper sum is !ˆ #i ‰ †
(d) ˜x œ
"! %
œ
" %
and xi œ i˜x œ
i %
Ê an upper sum is !ˆ 4i ‰ †
2. faxb œ x$
iœ! 2
iœ1 %
#
" )
#
#
#
#
" #
# œ "# Šˆ "# ‰ +1# ‹ œ
#
" 4
# # # œ 4" Šˆ 4" ‰ ˆ #" ‰ ˆ 4$ ‰ +1# ‹ œ
iœ"
" %
†
( )
" %
$! ‰ † ˆ "' œ
œ
( $#
& ) "& $#
Since f is increasing on Ò!ß "Ó, we use left endpoints to obtain lower sums and right endpoints to obtain upper sums.
"
$
iœ! $
" #
œ "# Š!$ ˆ "# ‰ ‹ œ
$
" 4
œ 4" Š!$ ˆ 4" ‰ ˆ #" ‰ ˆ 4$ ‰ ‹ œ
(a) ˜x œ
"! #
œ
" #
and xi œ i˜x œ
i #
Ê a lower sum is !ˆ #i ‰ †
(b) ˜x œ
"! %
œ
" %
and xi œ i˜x œ
i %
Ê a lower sum is !ˆ 4i ‰ †
(c) ˜x œ
"! #
œ
" #
and xi œ i˜x œ
i #
Ê an upper sum is !ˆ #i ‰ †
(d) ˜x œ
"! %
œ
" %
and xi œ i˜x œ
i %
Ê an upper sum is !ˆ 4i ‰ †
iœ! 2
iœ1 %
iœ"
$
" "'
$
$
$
$' #&'
$
" #
$ œ "# Šˆ "# ‰ +1$ ‹ œ
$
" 4
$ $ $ œ 4" Šˆ 4" ‰ ˆ #" ‰ ˆ 4$ ‰ +1$ ‹ œ œ
" #
†
* )
œ
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
œ
* '%
"!! #&'
œ
* "' #& '%
294
Chapter 5 Integration
3. faxb œ
" x
Since f is decreasing on Ò!ß "Ó, we use left endpoints to obtain upper sums and right endpoints to obtain lower sums.
#
(a) ˜x œ
&" #
œ # and xi œ " i˜x œ " #i Ê a lower sum is ! x"i † # œ #ˆ $" &" ‰ œ
(b) ˜x œ
&" %
œ 1 and xi œ " i˜x œ " i Ê a lower sum is
(c) ˜x œ
&" #
œ # and xi œ " i˜x œ " #i Ê an upper sum is ! x"i † # œ #ˆ" $" ‰ œ
(d) ˜x œ
&" %
œ 1 and xi œ " i˜x œ " i Ê an upper sum is
4. faxb œ % x#
iœ" % !" xi iœ" "
† " œ "ˆ #"
iœ! $ !" xi iœ!
" $
† " œ "ˆ"
" #
" %
"' "&
&" ‰ œ
" $
(( '!
) $
"% ‰ œ
#& "#
Since f is increasing on Ò#ß !Ó and decreasing on Ò!ß #Ó, we use left endpoints on Ò#ß !Ó and right endpoints on Ò!ß #Ó to obtain lower sums and use right endpoints on Ò#ß !Ó and left endpoints on Ò!ß #Ó to obtain upper sums.
(a) ˜x œ
# a#b #
œ # and xi œ # i˜x œ # #i Ê a lower sum is # † ˆ% a#b# ‰ # † a% ## b œ !
(b) ˜x œ
# a#b %
œ " and xi œ # i˜x œ # i Ê a lower sum is !ˆ% axi b# ‰ † " !ˆ% axi b# ‰ † "
"
%
iœ!
iœ$
œ "ˆˆ% a#b# ‰ ˆ% a"b# ‰ a% "# b a% ## b‰ œ ' (c) ˜x œ
# a#b #
œ # and xi œ # i˜x œ # #i Ê a upper sum is # † ˆ% a!b# ‰ # † a% !# b œ "'
(d) ˜x œ
# a#b %
œ " and xi œ # i˜x œ # i Ê a upper sum is !ˆ% axi b# ‰ † " !ˆ% axi b# ‰ † "
#
$
iœ"
iœ#
œ "ˆˆ% a"b# ‰ a% !# b a% !# b a% "# b‰ œ "% 5. faxb œ x#
œ
" #
Using 4 rectangles Ê ˜x œ " % ! œ Ê "% ˆfˆ ") ‰ fˆ $) ‰ fˆ &) ‰ fˆ () ‰‰
" %
Using 2 rectangles Ê ˜x œ #
#
œ "# Šˆ "% ‰ ˆ $% ‰ ‹ œ
#
#
"! $#
#
œ
"! #
Ê "# ˆfˆ "% ‰ fˆ $% ‰‰
& "'
#
œ "% Šˆ ") ‰ ˆ $) ‰ ˆ &) ‰ ˆ () ‰ ‹ œ
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
#" '%
Section 5.1 Estimating with Finite Sums 6. faxb œ x$
œ
" #
Using 4 rectangles Ê ˜x œ " % ! œ Ê "% ˆfˆ ") ‰ fˆ $) ‰ fˆ &) ‰ fˆ () ‰‰
" %
$
$
œ "# Šˆ "% ‰ ˆ $% ‰ ‹ œ
$
$
$
$
( œ "% Š " $ )& ‹œ $
7. faxb œ
" x
"! #
Using 2 rectangles Ê ˜x œ #) # † '%
%*' % † )$
œ
œ
Using 2 rectangles Ê ˜x œ œ #ˆ "# "% ‰ œ $#
Ê "# ˆfˆ "% ‰ fˆ $% ‰‰
( $#
"#% )$
&" #
œ
$" "#)
œ # Ê #afa#b fa%bb
Using 4 rectangles Ê ˜x œ & % " œ " Ê "ˆfˆ $# ‰ fˆ &# ‰ fˆ (# ‰ fˆ *# ‰‰ œ "ˆ #$
8. faxb œ % x#
# &
# (
#* ‰ œ
"%)) $†&†(†*
Using 2 rectangles Ê ˜x œ œ #a$ $b œ "#
295
œ
# a#b #
%*' &†(†*
œ
%*' $"&
œ # Ê #afa"b fa"bb
Using 4 rectangles Ê ˜x œ # %a#b œ " Ê "ˆfˆ $# ‰ fˆ "# ‰ fˆ "# ‰ fˆ $# ‰‰ # # # # œ "ŠŠ% ˆ $# ‰ ‹ Š% ˆ "# ‰ ‹ Š% ˆ "# ‰ ‹ Š% ˆ $# ‰ ‹‹ œ "' ˆ *% † # "% † #‰ œ "' "! # œ ""
9. (a) D ¸ (0)(1) (12)(1) (22)(1) (10)(1) (5)(1) (13)(1) (11)(1) (6)(1) (2)(1) (6)(1) œ 87 inches (b) D ¸ (12)(1) (22)(1) (10)(1) (5)(1) (13)(1) (11)(1) (6)(1) (2)(1) (6)(1) (0)(1) œ 87 inches 10. (a) D ¸ (1)(300) (1.2)(300) (1.7)(300) (2.0)(300) (1.8)(300) (1.6)(300) (1.4)(300) (1.2)(300) (1.0)(300) (1.8)(300) (1.5)(300) (1.2)(300) œ 5220 meters (NOTE: 5 minutes œ 300 seconds) (b) D ¸ (1.2)(300) (1.7)(300) (2.0)(300) (1.8)(300) (1.6)(300) (1.4)(300) (1.2)(300) (1.0)(300) (1.8)(300) (1.5)(300) (1.2)(300) (0)(300) œ 4920 meters (NOTE: 5 minutes œ 300 seconds) 11. (a) D ¸ (0)(10) (44)(10) (15)(10) (35)(10) (30)(10) (44)(10) (35)(10) (15)(10) (22)(10) (35)(10) (44)(10) (30)(10) œ 3490 feet ¸ 0.66 miles (b) D ¸ (44)(10) (15)(10) (35)(10) (30)(10) (44)(10) (35)(10) (15)(10) (22)(10) (35)(10) (44)(10) (30)(10) (35)(10) œ 3840 feet ¸ 0.73 miles 12. (a) The distance traveled will be the area under the curve. We will use the approximate velocities at the midpoints of each time interval to approximate this area using rectangles. Thus, D ¸ (20)(0.001) (50)(0.001) (72)(0.001) (90)(0.001) (102)(0.001) (112)(0.001) (120)(0.001) (128)(0.001) (134)(0.001) (139)(0.001) ¸ 0.967 miles (b) Roughly, after 0.0063 hours, the car would have gone 0.484 miles, where 0.0060 hours œ 22.7 sec. At 22.7 sec, the velocity was approximately 120 mi/hr.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
296
Chapter 5 Integration
13. (a) Because the acceleration is decreasing, an upper estimate is obtained using left end-points in summing acceleration † ?t. Thus, ?t œ 1 and speed ¸ [32.00 19.41 11.77 7.14 4.33](1) œ 74.65 ft/sec (b) Using right end-points we obtain a lower estimate: speed ¸ [19.41 11.77 7.14 4.33 2.63](1) œ 45.28 ft/sec (c) Upper estimates for the speed at each second are: t 0 1 2 3 4 5 v 0 32.00 51.41 63.18 70.32 74.65 Thus, the distance fallen when t œ 3 seconds is s ¸ [32.00 51.41 63.18](1) œ 146.59 ft. 14. (a) The speed is a decreasing function of time Ê right end-points give an lower estimate for the height (distance) attained. Also t 0 1 2 3 4 5 v 400 368 336 304 272 240 gives the time-velocity table by subtracting the constant g œ 32 from the speed at each time increment ?t œ 1 sec. Thus, the speed ¸ 240 ft/sec after 5 seconds. (b) A lower estimate for height attained is h ¸ [368 336 304 272 240](1) œ 1520 ft. 15. Partition [!ß #] into the four subintervals [0ß 0.5], [0.5ß 1], [1ß 1.5], and [1.5ß 2]. The midpoints of these subintervals are m" œ 0.25, m# œ 0.75, m$ œ 1.25, and m% œ 1.75. The heights of the four approximating 1 125 343 $ $ rectangles are f(m" ) œ (0.25)$ œ 64 , f(m# ) œ (0.75)$ œ 27 64 , f(m$ ) œ (1.25) œ 64 , and f(m% ) œ (1.75) œ 64 Notice that the average value is approximated by œ
" length of [!ß#]
†”
" #
$ $ $ $ ’ˆ 4" ‰ ˆ #" ‰ ˆ 34 ‰ ˆ #" ‰ ˆ 54 ‰ ˆ #" ‰ ˆ 74 ‰ ˆ #" ‰“ œ
$" "'
approximate area under • . We use this observation in solving the next several exercises. curve f(x) œ x$
16. Partition [1ß 9] into the four subintervals ["ß $], [3ß &], [&ß (], and [(ß *]. The midpoints of these subintervals are m" œ 2, m# œ 4, m$ œ 6, and m% œ 8. The heights of the four approximating rectangles are f(m" ) œ "# , f(m# ) œ "4 , f(m$ ) œ 6" , and f(m% ) œ 8" . The width of each rectangle is ?x œ 2. Thus, Area ¸ 2 ˆ "# ‰ 2 ˆ 4" ‰ 2 ˆ 6" ‰ 2 ˆ 8" ‰ œ
Ê average value ¸
25 1#
area length of ["ß*]
œ
ˆ 25 ‰ 12 8
œ
25 96 .
17. Partition [0ß 2] into the four subintervals [0ß 0.5], [0.5ß 1], [1ß 1.5], and [1.5ß 2]. The midpoints of the subintervals are m" œ 0.25, m# œ 0.75, m$ œ 1.25, and m% œ 1.75. The heights of the four approximating rectangles are " #
f(m" ) œ œ
" #
" #
sin#
1 4
œ
" #
" #
œ 1, and f(m% ) œ
œ 1, f(m# ) œ
" 2
sin#
71 4
œ
sin#
" #
Š È"2 ‹ œ 1. The width of each rectangle is ?x œ #" . Thus,
31 4
œ
" #
" #
œ 1, f(m$ ) œ
" 2
sin#
51 4
œ
" #
Š È"2 ‹
#
" 2
#
Area ¸ (1 1 1 1) ˆ "# ‰ œ 2 Ê average value ¸
area length of [0ß2]
œ
2 #
œ 1.
18. Partition [0ß 4] into the four subintervals [0ß 1], [1ß 2ß ], [2ß 3], and [3ß 4]. The midpoints of the subintervals are m" œ "# , m# œ #3 , m$ œ 5# , and m% œ 7# . The heights of the four approximating rectangles are f(m" ) œ 1 Šcos Š
% 1 ˆ "# ‰ 4 ‹‹
œ 1 ˆcos ˆ 18 ‰‰ œ 0.27145 (to 5 decimal places),
f(m# ) œ 1 Šcos Š
% 1 ˆ 3# ‰ 4 ‹‹
œ 1 ˆcos ˆ 381 ‰‰ œ 0.97855, f(m3 ) œ 1 Šcos Š
%
œ 0.97855, and f(m% ) œ 1 Šcos Š
%
% 1 ˆ 7# ‰ 4 ‹‹
% 1 ˆ #5 ‰ 4 ‹‹
œ 1 ˆcos ˆ 581 ‰‰
%
%
œ 1 ˆcos ˆ 781 ‰‰ œ 0.27145. The width of each rectangle is
?x œ ". Thus, Area ¸ (0.27145)(1) (0.97855)(1) (0.97855)(1) (0.27145)(1) œ 2.5 Ê average 2.5 5 value ¸ lengtharea of [0ß4] œ 4 œ 8 .
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 5.1 Estimating with Finite Sums
297
19. Since the leakage is increasing, an upper estimate uses right endpoints and a lower estimate uses left endpoints: (a) upper estimate œ (70)(1) (97)(1) (136)(1) (190)(1) (265)(1) œ 758 gal, lower estimate œ (50)(1) (70)(1) (97)(1) (136)(1) (190)(1) œ 543 gal. (b) upper estimate œ (70 97 136 190 265 369 516 720) œ 2363 gal, lower estimate œ (50 70 97 136 190 265 369 516) œ 1693 gal. (c) worst case: 2363 720t œ 25,000 Ê t ¸ 31.4 hrs; best case: 1693 720t œ 25,000 Ê t ¸ 32.4 hrs 20. Since the pollutant release increases over time, an upper estimate uses right endpoints and a lower estimate uses left endpoints: (a) upper estimate œ (0.2)(30) (0.25)(30) (0.27)(30) (0.34)(30) (0.45)(30) (0.52)(30) œ 60.9 tons lower estimate œ (0.05)(30) (0.2)(30) (0.25)(30) (0.27)(30) (0.34)(30) (0.45)(30) œ 46.8 tons (b) Using the lower (best case) estimate: 46.8 (0.52)(30) (0.63)(30) (0.70)(30) (0.81)(30) œ 126.6 tons, so near the end of September 125 tons of pollutants will have been released. #
21. (a) The diagonal of the square has length 2, so the side length is È#. Area œ ŠÈ#‹ œ # (b) Think of the octagon as a collection of 16 right triangles with a hypotenuse of length 1 and an acute angle measuring #1 1 "' œ ) . Area œ "'ˆ " ‰ˆsin 1 ‰ˆcos 1 ‰ œ % sin 1 œ #È# ¸ #Þ)#) #
)
)
%
(c) Think of the 16-gon as a collection of 32 right triangles with a hypotenuse of length 1 and an acute angle measuring #1 1 $# œ "' . Area œ $#ˆ " ‰ˆsin 1 ‰ˆcos 1 ‰ œ ) sin 1 œ #È# ¸ $Þ!'" #
"'
"'
)
(d) Each area is less than the area of the circle, 1. As n increases, the area approaches 1. 22. (a) Each of the isosceles triangles is made up of two right triangles having hypotenuse 1 and an acute angle measuring #1 1 1 ‰ˆ ˆ " ‰ˆ cos 1n ‰ œ "# sin #n1 . #n œ n . The area of each isosceles triangle is AT œ # # sin n (b) The area of the polygon is AP œ nAT œ
n #
sin
#1 n ,
n nÄ_ #
so lim
sin
#1 n
œ lim 1 † nÄ_
sin #n1 ˆ #n1 ‰
œ1
(c) Multiply each area by r# . AT œ "# r# sin #n1 AP œ n# r# sin lim AP œ 1r
#
#1 n
nÄ_
23-26. Example CAS commands: Maple: with( Student[Calculus1] ); f := x -> sin(x); a := 0; b := Pi; plot( f(x), x=a..b, title="#23(a) (Section 5.1)" ); N := [ 100, 200, 1000 ]; # (b) for n in N do Xlist := [ a+1.*(b-a)/n*i $ i=0..n ]; Ylist := map( f, Xlist ); end do: for n in N do # (c) Avg[n] := evalf(add(y,y=Ylist)/nops(Ylist)); Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
298
Chapter 5 Integration
end do; avg := FunctionAverage( f(x), x=a..b, output=value ); evalf( avg ); FunctionAverage(f(x),x=a..b,output=plot); # (d) fsolve( f(x)=avg, x=0.5 ); fsolve( f(x)=avg, x=2.5 ); fsolve( f(x)=Avg[1000], x=0.5 ); fsolve( f(x)=Avg[1000], x=2.5 ); Mathematica: (assigned function and values for a and b may vary): Symbols for 1, Ä , powers, roots, fractions, etc. are available in Palettes (under File). Never insert a space between the name of a function and its argument. Clear[x] f[x_]:=x Sin[1/x] {a,b}={1/4, 1} Plot[f[x],{x, a, b}] The following code computes the value of the function for each interval midpoint and then finds the average. Each sequence of commands for a different value of n (number of subdivisions) should be placed in a separate cell. n =100; dx = (b a) /n; values = Table[N[f[x]], {x, a dx/2, b, dx}] average=Sum[values[[i]],{i, 1, Length[values]}] / n n =200; dx = (b a) /n; values = Table[N[f[x]],{x, a + dx/2, b, dx}] average=Sum[values[[i]],{i, 1, Length[values]}] / n n =1000; dx = (b a) /n; values = Table[N[f[x]],{x, a dx/2, b, dx}] average=Sum[values[[i]],{i, 1, Length[values]}] / n FindRoot[f[x] == average,{x, a}] 5.2 SIGMA NOTATION AND LIMITS OF FINITE SUMS 2
1. ! kœ1
3
2. ! kœ1
6k k1
œ
6(1) 11
6(2) 21
œ
6 2
k1 k
œ
11 1
21 2
31 3
12 3
œ7
œ0
1 2
2 3
œ
7 6
4
3. ! cos k1 œ cos (11) cos (21) cos (31) cos (41) œ 1 1 1 1 œ 0 kœ1
5
4. ! sin k1 œ sin (11) sin (21) sin (31) sin (41) sin (51) œ 0 0 0 0 0 œ 0 kœ1
3
5. ! (1)kb1 sin kœ1
1 k
œ (1)"" sin
1 1
(1)#" sin
1 #
(")$" sin
1 3
œ 01
È3 #
œ
È3 2 #
4
6. ! (1)k cos k1 œ (1)" cos (11) (1)# cos (21) (1)$ cos (31) (1)% cos (41) kœ1
œ (1) 1 (1) 1 œ 4
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 5.2 Sigma Notation and Limits of Finite Sums 6
7. (a) ! 2kc1 œ 2"" 2#" 2$" 2%" 2&" 2'" œ 1 2 4 8 16 32 kœ1 5
(b) ! 2k œ 2! 2" 2# 2$ 2% 2& œ 1 2 4 8 16 32 kœ0 4
(c) ! 2k1 œ 2"" 2!" 2"" 2#" 2$" 2%" œ 1 2 4 8 16 32 kœ"
All of them represent 1 2 4 8 16 32 6
8. (a) ! (2)k1 œ (2)"" (2)#" (2)$" (2)%" (2)&" (2)'" œ 1 2 4 8 16 32 kœ1 5
(b) ! (1)k 2k œ (1)! 2! Ð")" 2" (1)# 2# (1)$ 2$ (1)% 2% (1)& 2& œ 1 2 4 8 16 32 kœ0 3
(c) ! (1)k1 2k2 œ Ð")#" 2## (")"" 2"# (")!" 2!# (1)"" 2"# (")#" 2## kœ2
(1)$" 2$# œ 1 2 4 8 16 32; (a) and (b) represent 1 2 4 8 16 32; (c) is not equivalent to the other two 4
9. (a) ! kœ2 2
(b) ! kœ0 1
(c) !
kœ"
(")k" k1
(1)#" 21
œ
(")k k1
œ
(1)! 01
(")k k2
œ
(1)" 1 2
(")$" 31
(")" 11
(")! 02
(")# 21
(")%" 41
œ 1
œ1
(")" 12
" #
œ 1
" #
" #
" 3
" 3
" 3
(a) and (c) are equivalent; (b) is not equivalent to the other two. 4
10. (a) ! (k 1)# œ (1 1)# (2 1)# (3 1)# (4 1)# œ 0 1 4 9 kœ1 3
(b) ! (k 1)# œ (1 1)# (0 1)# (1 1)# (2 1)# (3 1)# œ 0 1 4 9 16 kœ1
"
(c) ! k# œ (3)# (2)# (1)# œ 9 4 1 kœ3
(a) and (c) are equivalent to each other; (b) is not equivalent to the other two. 6
4
4
12. ! k#
11. ! k kœ1
13. !
kœ1
5
5
15. ! (1)k1
14. ! 2k kœ1
kœ1
n
kœ1
" k
5
16. ! (1)k kœ1
n
17. (a) ! 3ak œ 3 ! ak œ 3(5) œ 15 kœ1 n
(b) ! kœ1 n
bk 6
œ
" 6
kœ1 n
! bk œ kœ1
" 6
(6) œ 1
n
n
kœ1 n
kœ1 n
kœ1 n
kœ1 n
kœ1 n
kœ1
(c) ! (ak bk ) œ ! ak ! bk œ 5 6 œ 1 (d) ! (ak bk ) œ ! ak ! bk œ 5 6 œ 11 n
(e) ! (bk 2ak ) œ ! bk 2 ! ak œ 6 2(5) œ 16 kœ1
kœ1
" #k
kœ1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
k 5
299
300
Chapter 5 Integration n
n
kœ1 n
kœ1
n
18. (a) ! 8ak œ 8 ! ak œ 8(0) œ 0 n
n
kœ1
kœ1
(c) ! (ak 1) œ ! ak ! 1 œ 0 n œ n kœ1
10
19. (a) ! k œ kœ1
10(10 1) #
n
(b) ! 250bk œ 250 ! bk œ 250(1) œ 250 kœ1 n
n
kœ1
n
kœ1
kœ1
kœ1
(d) ! (bk 1) œ ! bk ! 1 œ " n 10
(b) ! k# œ
œ 55
kœ1
10(10 1)(2(10) 1) 6
œ 385
13(13 1)(2(13) 1) 6
œ 819
#
10
(c) ! k$ œ ’ 10(10# 1) “ œ 55# œ 3025 kœ1
13
20. (a) ! k œ kœ1
13(13 1) #
13
(b) ! k# œ
œ 91
kœ1
#
13
(c) ! k$ œ ’ 13(13# 1) “ œ 91# œ 8281 kœ1
7
7
kœ1
kœ1
6
6
6
kœ1
kœ1
kœ1
6
6
6
kœ1
kœ1
kœ1
5
21. ! 2k œ 2 ! k œ 2 Š 7(7 # ") ‹ œ 56
23. ! a3 k# b œ ! 3 ! k# œ 3(6)
24. ! ak# 5b œ ! k# ! 5 œ
22. ! kœ1
6(6 ")(2(6) 1) 6
6(6 ")(2(6) 1) 6
1k 15
œ
1 15
5
!kœ kœ1
1 15
Š 5(5 # 1) ‹ œ 1
œ 73
5(6) œ 61
5
5
5
5
kœ1
kœ1
kœ1
kœ1
7
7
7
7
kœ1
kœ1
kœ1
kœ1
1) 25. ! k(3k 5) œ ! a3k# 5kb œ 3 ! k# 5 ! k œ 3 Š 5(5 1)(2(5) ‹ 5 Š 5(5 # 1) ‹ œ 240 6
1) 26. ! k(2k 1) œ ! a2k# kb œ 2 ! k# ! k œ 2 Š 7(7 1)(2(7) ‹ 6
5
k$ 225
27. ! kœ1
7
kœ1
#
7
28. Œ! k ! kœ1
29. (a)
$
5
Œ! k œ
kœ1
k$ 4
" 2 #5
7
5
5
kœ1
kœ1
#
œ Œ! k kœ1
$
! k $ Œ! k œ
" 4
" #25
#
! k$ œ Š 7(7 1) ‹ # kœ1
(b)
œ 308
$
Š 5(5 # 1) ‹ Š 5(5 # 1) ‹ œ 3376 #
7
7(7 1) #
" 4
#
Š 7(7 # 1) ‹ œ 588 (c)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 5.2 Sigma Notation and Limits of Finite Sums 30. (a)
(b)
(c)
31. (a)
(b)
(c)
32. (a)
(b)
(c)
301
33. kx" x! k œ k1.2 0k œ 1.2, kx# x" k œ k1.5 1.2k œ 0.3, kx$ x# k œ k2.3 1.5k œ 0.8, kx% x$ k œ k2.6 2.3k œ 0.3, and kx& x% k œ k3 2.6k œ 0.4; the largest is lPl œ 1.2. 34. kx" x! k œ k1.6 (2)k œ 0.4, kx# x" k œ k0.5 (1.6)k œ 1.1, kx$ x# k œ k0 (0.5)k œ 0.5, kx% x$ k œ k0.8 0k œ 0.8, and kx& x% k œ k1 0.8k œ 0.2; the largest is lPl œ 1.1. 35. faxb œ " x#
Since f is decreasing on Ò !, 1Ó we use left endpoints to obtain upper sums. ˜x œ " n ! œ "n and xi œ i˜x œ ni . So an upper sum n"
is ! a" x#i b "n œ iœ!
œ
n$ n$
n "! # i n$ iœ!
" n
n"
! Š" ˆ i ‰# ‹ œ n
iœ!
œ"
an " b n a# an " b " b 'n $
# $n n"# . Thus, ' n" lim ! a" x#i b "n œ lim Œ" nÄ_ iœ! nÄ_
" n$
n"
! an# i# b
iœ!
œ"
#n$ $n# n 'n $
œ"
# $n n"# '
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
œ"
" $
œ
# $
302
Chapter 5 Integration Since f is increasing on Ò !, $Ó we use right endpoints to obtain upper sums. ˜x œ $ n ! œ $n and xi œ i˜x œ $ni . So an upper
36. faxb œ #x
n
n
sum is !#xi ˆ n$ ‰ œ ! 'ni † iœ"
iœ"
n
Thus,
37. faxb œ x# "
lim ! 'i nÄ_ iœ" n
†
$ n
œ
# œ lim *n n# *n nÄ_
$ n
n
") n#
") n#
!i œ
iœ"
n an " b #
†
œ
*n# *n n#
œ lim ˆ* n* ‰ œ *. nÄ_
Since f is increasing on Ò !, $Ó we use right endpoints to obtain upper sums. ˜x œ $ n ! œ $n and xi œ i˜x œ $ni . So an upper n
n
# sum is !ax#i "b $n œ !Šˆ $ni ‰ "‹ n$ œ
œ
iœ" n #( ! # i n$ n iœ"
iœ"
†nœ
#( nan "ba#n "b ‹ n$ Š '
n
!Š *i## "‹ n
$ n
iœ"
$
* ") #( *a#n$ $n# nb n n# $œ $Þ Thus, #n $ # n #( ") n n*# lim !ax#i "b $n œ lim Œ $ œ # nÄ_ iœ" nÄ_
œ
38. faxb œ $x#
Since f is increasing on Ò !, "Ó we use right endpoints to obtain upper sums. ˜x œ " n ! œ "n and xi œ i˜x œ ni . So an upper sum n
n
iœ"
iœ"
# is !$x#i ˆ "n ‰ œ !$ˆ ni ‰ ˆ n" ‰ œ
œ
#n$ $n# n #n $
œ lim Œ nÄ_
39. faxb œ x x# œ xa" xb
œ
# $n n"# #
# $n n"# # # #
œ
n
$ n$
! i# œ
iœ"
$ n$
† Š nan "ba' #n "b ‹
n
. Thus, lim !$x#i ˆ "n ‰ nÄ_ iœ"
œ ".
Since f is increasing on Ò !, "Ó we use right endpoints to obtain upper sums. ˜x œ " n ! œ "n and xi œ i˜x œ ni . So an upper sum n
n
# is !axi xi# b n" œ !Š ni ˆ ni ‰ ‹ n" œ iœ"
iœ"
œ
" n a n "b ‹ n# Š #
œ
" "n #
n"$ Š nan "ba' #n "b ‹ n # $ "
œ lim ”Š nÄ_
40. faxb œ $x #x#
* $ œ "#.
n
'
" #
" n
n#
n "! i n# iœ"
n "! # i n$ iœ"
n# n #n#
#n $ $ n # n 'n$
œ
. Thus, lim !axi x#i b "n nÄ_ iœ"
‹Œ
$ n
# n"# '
• œ
" #
# '
œ &' .
Since f is increasing on Ò !, "Ó we use right endpoints to obtain upper sums. ˜x œ " n ! œ "n and xi œ i˜x œ ni . So an upper sum n
n
# is !a$xi #x#i b "n œ !Š $ni #ˆ ni ‰ ‹ n" œ iœ"
iœ"
œ
$ n a n "b ‹ n# Š #
œ
$ $n #
œ lim ”Š nÄ_
n#$ Š nan "ba' #n "b ‹ n # $ "
$ #
n
$
$ n
n#
œ
n $! i n# iœ"
$n# $n #n#
. Thus, lim !a$xi #x#i b "n
‹Œ
nÄ_ iœ"
$ n
# n"# $
• œ
$ #
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
# $
œ
"$ ' .
n #! # i n$ iœ"
#n# $n " $n#
Section 5.3 The Definite Integral
303
5.3 THE DEFINITE INTEGRAL
1.
'02 x# dx
2.
'"! 2x$ dx
3.
'(& ax# 3xb dx
4.
'"% "x dx
5.
'#$ 1 " x dx
6.
'0" È4 x# dx
7.
'! Î% (sec x) dx
8.
'0 Î% (tan x) dx 1
1
9. (a) (c) (e) (f)
10. (a) (b) (c) (d) (e) (f)
11. (a) (c)
12. (a) (c)
13. (a) (b)
14. (a) (b)
" & '#2 g(x) dx œ 0 (b) ' g(x) dx œ ' g(x) dx œ 8 & " 2 2 & & 2 '" 3f(x) dx œ 3'" f(x) dx œ 3(4) œ 12 (d) ' f(x) dx œ ' f(x) dx ' f(x) dx œ 6 (4) œ 10 # " " & & & '" [f(x) g(x)] dx œ '" f(x) dx '" g(x) dx œ 6 8 œ 2 '"& [4f(x) g(x)] dx œ 4 '"& f(x) dx '"& g(x) dx œ 4(6) 8 œ 16
'"* 2f(x) dx œ 2 '"* f(x) dx œ 2(1) œ 2 '(* [f(x) h(x)] dx œ '(*f(x) dx '(* h(x) dx œ 5 4 œ 9 '(* [2f(x) 3h(x)] dx œ 2 '(* f(x) dx 3 '(* h(x) dx œ 2(5) 3(4) œ 2 '*"f(x) dx œ '"* f(x) dx œ (1) œ 1 '"( f(x) dx œ '"* f(x) dx '(* f(x) dx œ 1 5 œ 6 '*( [h(x) f(x)] dx œ '(* [f(x) h(x)] dx œ '(* f(x) dx '(* h(x) dx œ 5 4 œ 1 '"2 f(u) du œ '"2 f(x) dx œ 5 '#" f(t) dt œ '"2 f(t) dt œ 5 '!$ g(t) dt œ '$! g(t) dt œ È2 '$! [g(x)] dx œ '$! g(x) dx œ È2
(b) (d)
(b) (d)
'"2 È3 f(z) dz œ È3 '"2 f(z) dz œ 5È3 '"2 [f(x)] dx œ '"2 f(x) dx œ 5 '$! g(u) du œ '$! g(t) dt œ È2 '$! Èg(r)2 dr œ È"2 '$! g(t) dt œ Š È"2 ‹ ŠÈ2‹ œ 1
'$% f(z) dz œ '!% f(z) dz '!$ f(z) dz œ 7 3 œ 4 '%$ f(t) dt œ '$% f(t) dt œ 4 '"$ h(r) dr œ '"$ h(r) dr '"" h(r) dr œ 6 0 œ 6 " $ $ ' h(u) du œ Œ ' h(u) du œ ' h(u) du œ 6 $ " "
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
304
Chapter 5 Integration
15. The area of the trapezoid is A œ œ
" #
(5 2)(6) œ 21 Ê
œ 21 square units
" #
(3 1)(1) œ 2 Ê
(B b)h
" #
(B b)h
'# ˆ #x 3‰ dx %
16. The area of the trapezoid is A œ œ
" #
'"Î#
$Î#
(2x 4) dx
œ 2 square units
17. The area of the semicircle is A œ œ
9 #
1 Ê
" #
1r# œ
1(3)#
'$$ È9 x# dx œ 9# 1 square units
18. The graph of the quarter circle is A œ œ 41 Ê
" #
" 4
1 r# œ
" 4
1(4)#
'%! È16 x# dx œ 41 square units
19. The area of the triangle on the left is A œ
" #
bh œ
œ 2. The area of the triangle on the right is A œ œ Ê
" #
(1)(1) œ
" #.
" # " #
(2)(2) bh
Then, the total area is 2.5
'# kxk dx œ 2.5 square units "
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 5.3 The Definite Integral 20. The area of the triangle is A œ Ê
" #
bh œ
'" a1 kxkb dx œ 1 square unit "
21. The area of the triangular peak is A œ
" #
(2)(1) œ 1
" #
bh œ
" #
(2)(1) œ 1.
The area of the rectangular base is S œ jw œ (2)(1) œ 2. Then the total area is 3 Ê
'"" a2 kxkb dx œ 3 square units
22. y œ 1 È1 x# Ê y 1 œ È1 x# Ê (y 1)# œ 1 x# Ê x# (y 1)# œ 1, a circle with center (!ß ") and radius of 1 Ê y œ 1 È1 x# is the upper semicircle. The area of this semicircle is A œ "# 1r# œ "# 1(1)# œ 1# . The area of the rectangular base is A œ jw œ (2)(1) œ 2. Then the total area is 2 Ê
23.
'"" Š1 È1 x# ‹ dx œ 2 1# square units
'!b x2 dx œ "# (b)( b2 ) œ b4
#
1 #
24.
'!b 4x dx œ "# b(4b) œ 2b#
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305
306
Chapter 5 Integration
25.
'ab 2s ds œ "# b(2b) "# a(2a) œ b# a#
27.
'"
29.
'1#1 ) d) œ (2#1)
31.
'0
33.
'!"Î# t# dt œ ˆ 3‰
œ
35.
'a#a x dx œ (2a)#
37.
'!
39.
'$" 7 dx œ 7(1 3) œ 14
41.
'!2 5x dx œ 5 '!2 x dx œ 5 ’ 2#
43.
'!2 (2t 3) dt œ 2 '"" t dt '!2 3 dt œ 2 ’ 2#
44.
'!
45.
'#" ˆ1 #z ‰ dz œ '#" 1 dz '#" #z dz œ '#" 1 dz "# '"# z dz œ 1[1 2] "# ’ 2# 1# “ œ " "# ˆ 3# ‰ œ 74
46.
'$! (2z 3) dz œ '$! 2z dz '$! 3 dz œ 2 '!$ z dz '$! 3 dz œ 2 ’ 3#
47.
'"# 3u# du œ 3 '"# u# du œ 3 ”'!# u# du '!" u# du• œ 3 Š’ 23
È#
3 È 7
ŠÈ2‹ #
#
(1)# #
œ
1# #
œ
31 # #
" #
28.
'!Þ&#Þ& x dx œ (2.5)#
30.
'È& # # r dr œ Š5È#2‹
x dx œ
3 7‹ ŠÈ
œ
32.
'!!Þ$ s# ds œ (0.3)3
3
34.
'!1Î# )# d) œ ˆ 3‰
7 3
" 24
a# #
œ
36.
'a
$ b‹ ŠÈ
3
œ
b 3
38.
'!$b x# dx œ (3b)3
40.
'!2 È2 dx œ È2 (# !) œ 2È2
#
0# #“
œ 10
42.
'$& 8x dx œ "8 '$& x dx œ 8" ’ 5#
#
Št È2‹ dt œ
È2
'!
È2
t dt ' !
0# #“
(0.5)# #
#
$
1 $ #
È$a
3a# #
$
x# dx œ
#
È
$
#
#
È2
'ab 3t dt œ "# b(3b) "# a(3a) œ 3# ab# a# b
#
x dx œ
" $ #
$ È b
26.
œ3 #
ŠÈ2‹ #
œ 24
œ 0.009
œ
1$ #4
#
ŠÈ3a‹
x dx œ
#
$
a# #
œ a#
œ 9b$
#
3# #“
œ
16 16
œ1
3(2 0) œ 4 6 œ 2 #
È2 dt œ
ŠÈ2‹
–
#
0# #—
È2 ’È2 0“ œ 1 2 œ 1
#
#
$
0$ 3“
$
’ "3
0# #“
#
3[0 3] œ 9 9 œ 0
0$ 3 “‹
$
œ 3 ’ 23
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
1$ 3“
œ 3 ˆ 73 ‰ œ 7
Section 5.3 The Definite Integral 48.
'"Î#" 24u# du œ 24 '"Î#"
49.
'!# a3x# x 5b dx œ 3 '!# x# dx '!# x dx '!# 5 dx œ 3 ’ 23
50.
'"! a3x# x 5b dx œ '!" a3x# x 5b dx œ ”3 '!" x# dx '!" x dx '!" 5 dx•
u# du œ 24 –
'!"
u# du
'!"Î#
$
u# du— œ 24 ” 13
$
$
0$ 3‹
b0 n
œ
œ ’3 Š 13 51. Let ?x œ
#
Š 1#
0# #‹
5(1 0)“ œ ˆ 3# 5‰ œ
0$ 3“
ˆ "# ‰$ 3 •
#
’ 2#
œ 24 ’
0# #“
ˆ 78 ‰ 3
“œ7
5[2 0] œ (8 2) 10 œ 0
7 #
and let x! œ 0, x" œ ?x,
b n
x# œ 2?xß á ß xn" œ (n 1)?x, xn œ n?x œ b. Let the ck 's be the right end-points of the subintervals Ê c" œ x" , c# œ x# , and so on. The rectangles defined have areas: f(c" ) ?x œ f(?x) ?x œ 3(?x)# ?x œ 3(?x)$ f(c# ) ?x œ f(2?x) ?x œ 3(2?x)# ?x œ 3(2)# (?x)$ f(c$ ) ?x œ f(3?x) ?x œ 3(3?x)# ?x œ 3(3)# (?x)$ ã f(cn ) ?x œ f(n?x) ?x œ 3(n?x)# ?x œ 3(n)# (?x)$ n
n
kœ1 n
kœ1
Then Sn œ ! f(ck ) ?x œ ! 3k# (?x)$ $ 1) œ 3(?x) ! k# œ 3 Š bn$ ‹ Š n(n 1)(2n ‹ 6
$
kœ1
œ
$
b #
ˆ2
52. Let ?x œ
3 n
b0 n
"‰ n#
œ
Ê
b n
'!b 3x# dx œ n lim Ä_
b$ #
ˆ2
3 n
"‰ n#
œ b$ .
and let x! œ 0, x" œ ?x,
x# œ 2?xß á ß xn" œ (n 1)?x, xn œ n?x œ b. Let the ck 's be the right end-points of the subintervals Ê c" œ x" , c# œ x# , and so on. The rectangles defined have areas: f(c" ) ?x œ f(?x) ?x œ 1(?x)# ?x œ 1(?x)$ f(c# ) ?x œ f(2?x) ?x œ 1(2?x)# ?x œ 1(2)# (?x)$ f(c$ ) ?x œ f(3?x) ?x œ 1(3?x)# ?x œ 1(3)# (?x)$ ã f(cn ) ?x œ f(n?x) ?x œ 1(n?x)# ?x œ 1(n)# (?x)$ n
n
Then Sn œ ! f(ck ) ?x œ ! 1k# (?x)$ kœ1 n
kœ1
1) œ 1(?x)$ ! k# œ 1 Š bn$ ‹ Š n(n 1)(2n ‹ 6 $
kœ1
œ
1b 6
$
ˆ2
3 n
"‰ n#
Ê
'!b 1x# dx œ n lim Ä_
1 b$ 6
ˆ2
3 n
"‰ n#
œ
307
1 b$ 3 .
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308
Chapter 5 Integration b0 n
53. Let ?x œ
œ
b n
and let x! œ 0, x" œ ?x,
x# œ 2?xß á ß xn" œ (n 1)?x, xn œ n?x œ b. Let the ck 's be the right end-points of the subintervals Ê c" œ x" , c# œ x# , and so on. The rectangles defined have areas: f(c" ) ?x œ f(?x) ?x œ 2(?x)(?x) œ 2(?x)# f(c# ) ?x œ f(2?x) ?x œ 2(2?x)(?x) œ 2(2)(?x)# f(c$ ) ?x œ f(3?x) ?x œ 2(3?x)(?x) œ 2(3)(?x)# ã f(cn ) ?x œ f(n?x) ?x œ 2(n?x)(?x) œ 2(n)(?x)# n
n
Then Sn œ ! f(ck ) ?x œ ! 2k(?x)# kœ1 n
kœ1
œ 2(?x)# ! k œ 2 Š bn# ‹ Š n(n 2 1) ‹ #
kœ1
œ b# ˆ1 "n ‰ Ê b0 n
54. Let ?x œ
œ
'!b 2x dx œ n lim Ä_ b n
b# ˆ1 n" ‰ œ b# .
and let x! œ 0, x" œ ?x,
x# œ 2?xß á ß xn" œ (n 1)?x, xn œ n?x œ b. Let the ck 's be the right end-points of the subintervals Ê c" œ x" , c# œ x# , and so on. The rectangles defined have areas: " # ‰ f(c" ) ?x œ f(?x) ?x œ ˆ ?x # 1 (?x) œ # (?x) ?x 2 ? x " f(c# ) ?x œ f(2?x) ?x œ ˆ # 1‰ (?x) œ # (2)(?x)# ?x f(c$ ) ?x œ f(3?x) ?x œ ˆ 3?# x 1‰ (?x) œ
" #
(3)(?x)# ?x
f(cn ) ?x œ f(n?x) ?x œ ˆ n?# x 1‰ (?x) œ
" #
(n)(?x)# ?x
ã
n
n
kœ1
kœ1
Then Sn œ ! f(ck ) ?x œ ! ˆ "# k(?x)# ?x‰ œ œ
" 4
b# ˆ1 n1 ‰ b Ê
55. av(f) œ Š È3" 0 ‹ œ
'! ˆ x# 1‰ dx œ n lim Ä_ b
" È3
È$
'!
È$
'!
x# dx
" #
n
n
kœ1
kœ1
(?x)# ! k ?x ! 1 œ
ˆ 4" b# ˆ1 n" ‰ b‰ œ
" 4
" #
Š bn# ‹ Š n(n 2 1) ‹ ˆ bn ‰ (n) #
b# b.
ax# 1b dx È$
'!
" È3
1 dx
$
œ
" È3
ŠÈ3‹
3
56. av(f) œ ˆ 3 " 0 ‰ $
" È3
ŠÈ3 0‹ œ 1 1 œ 0.
'!$ Š x# ‹ dx œ 3" ˆ #" ‰ '!$ x# dx #
#
œ "6 Š 33 ‹ œ 3# ; x# œ 3# .
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Section 5.3 The Definite Integral
'!" a3x# 1b dx œ " " œ 3 ' x# dx ' 1 dx œ 3 Š 13 ‹ (1 0) ! !
57. av(f) œ ˆ 1 " 0 ‰
$
œ #.
'!" a3x# 3b dx œ " " œ 3 ' x# dx ' 3 dx œ 3 Š 13 ‹ 3(1 0) ! !
58. av(f) œ ˆ 1 " 0 ‰
$
œ #.
'!$ (t 1)# dt $ $ $ œ 3" ' t# dt 23 ' t dt 3" ' 1 dt ! ! !
59. av(f) œ ˆ 3 " 0 ‰
œ
" 3
$
#
Š 33 ‹ 32 Š 3#
60. av(f) œ Š 1 1(2) ‹
0# #‹
3" (3 0) œ 1.
'#" at# tb dt
'#" t# dt 3" '#" t dt " # œ "3 ' t# dt 3" ' t# dt 3" Š 1# ! ! œ
" 3
#
œ
" 3
$
Š 13 ‹ 3" Š (32) ‹ $
61. (a) av(g) œ Š 1 "(1) ‹
" #
œ
3 #
(2)# # ‹
.
'"" akxk 1b dx
'"! (x 1) dx "# '!" (x 1) dx ! ! " " œ "# ' x dx "# ' 1 dx "# ' x dx "# ' 1 dx " " ! ! œ
" #
#
œ "# Š 0#
(1)# # ‹
#
"# (0 (1)) "# Š 1#
0# #‹
"# (1 0)
œ "# .
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309
310
Chapter 5 Integration
'"$ akxk 1b dx œ #" '"$ (x 1) dx $ $ œ "# ' x dx "# ' 1 dx œ "# Š 3# 1# ‹ "# (3 1) " "
(b) av(g) œ ˆ 3 " 1 ‰
#
#
œ 1.
(c) av(g) œ Š 3 "(1) ‹ œ
" 4 " 4
'"$ akxk 1b dx
'"" akxk 1b dx 4" '"$ akxk 1b dx " 4
(see parts (a) and (b) above).
62. (a) av(h) œ Š 0 "(1) ‹
'"0 kxk dx œ '"0 (x) dx
œ
œ
(1 2) œ
'"0 x dx œ 0#
#
(b) av(h) œ ˆ 1 " 0 ‰ #
œ Š "#
(1)# #
œ "# .
'0" kxk dx œ '0" x dx
0# #‹
œ "# .
(c) av(h) œ Š 1 "(1) ‹
'"" kxk dx
'"0 kxk dx '0" kxk dx
œ
" #
Œ
œ
" #
ˆ "# ˆ "# ‰‰ œ "# (see parts (a) and (b)
above).
63. To find where x x# 0, let x x# œ 0 Ê x(1 x) œ 0 Ê x œ 0 or x œ 1. If 0 x 1, then 0 x x# Ê a œ 0 and b œ 1 maximize the integral.
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Section 5.3 The Definite Integral
311
64. To find where x% 2x# Ÿ 0, let x% 2x# œ 0 Ê x# ax# 2b œ 0 Ê x œ 0 or x œ „ È2. By the sign graph, 0 0 0 , we can see that x% 2x# Ÿ 0 on ’È2ß È2“ Ê a œ È2 and b œ È2 ! È# È # minimize the integral. " 1 x #
65. f(x) œ
is decreasing on [0ß 1] Ê maximum value of f occurs at 0 Ê max f œ f(0) œ 1; minimum value of f
occurs at 1 Ê min f œ f(1) œ Ê
" #
Ÿ
'0" 1 " x
" 1 1#
œ
" #
. Therefore, (1 0) min f Ÿ
dx Ÿ 1. That is, an upper bound œ 1 and a lower bound œ
#
66. See Exercise 65 above. On [0ß 0.5], max f œ
'0
0.5
(0.5 0) min f Ÿ min f œ Then
" 4
" 1 1#
2 5
" 1 0#
'0
0.5
" 1 x#
dx
'0.5 1 " x "
#
dx Ÿ
67. 1 Ÿ sin ax# b Ÿ 1 for all x Ê (1 0)(1) Ÿ
" #
" 1 (0.5)#
œ 1, min f œ
f(x) dx Ÿ (0.5 0) max f Ê
Ÿ
2 5
'0
'0.5 1 " x "
œ 0.5. Therefore (1 0.5) min f Ÿ
Ÿ
'0" 1 " x
2 5
Ê
0.5
#
" 1 x#
#
" #
dx Ÿ (1 0) max f .
œ 0.8. Therefore
dx Ÿ
" #
. On [0.5ß 1], max f œ
dx Ÿ (1 0.5) max f Ê
13 20
Ÿ
'0 1 " x "
#
dx Ÿ
9 10
" 4
Ÿ
" 1 (0.5)#
'0.5 1 1 x "
#
dx Ÿ
œ 0.8 and 2 5
.
.
'0" sin ax# b dx Ÿ (1 0)(1) or '0"sin x# dx Ÿ 1
Ê
'0"sin x# dx cannot
equal 2. 68. f(x) œ Èx 8 is increasing on [!ß "] Ê max f œ f(1) œ È1 8 œ 3 and min f œ f(0) œ È0 8 œ 2È2 . Therefore, (1 0) min f Ÿ
'0" Èx 8 dx Ÿ (1 0) max f
Ê 2È 2 Ÿ
'0" Èx 8 dx Ÿ 3.
69. If f(x) 0 on [aß b], then min f 0 and max f 0 on [aß b]. Now, (b a) min f Ÿ Then b a Ê b a 0 Ê (b a) min f 0 Ê
'ab f(x) dx 0.
70. If f(x) Ÿ 0 on [aß b], then min f Ÿ 0 and max f Ÿ 0. Now, (b a) min f Ÿ b a Ê b a 0 Ê (b a) max f Ÿ 0 Ê
'ab f(x) dx Ÿ 0.
'ab f(x) dx Ÿ (b a) max f.
'ab f(x) dx Ÿ (b a) max f.
Then
'0" (sin x x) dx Ÿ 0 (see Exercise 70) Ê '0" sin x dx '0" x dx '0" sin x dx Ÿ Š 1# 0# ‹ Ê '0" sin x dx Ÿ "# . Thus an upper bound is "# .
71. sin x Ÿ x for x 0 Ê sin x x Ÿ 0 for x 0 Ê Ÿ0 Ê
'0" sin x dx Ÿ '0" x dx
72. sec x 1
x# #
Ê
#
on ˆ 1# ß 1# ‰ Ê sec x Š1
x# #‹
0 on ˆ 1# ß 1# ‰ Ê
Exercise 69) since [0ß 1] is contained in ˆ 1# ß 1# ‰ Ê
'0" Š1 x# ‹ dx #
Ê
#
'0" ’sec x Š1 x# ‹“ dx 0 (see #
'0"sec x dx '0" Š1 x# ‹ dx 0
'0" sec x dx '0" 1 dx "# '0" x# dx
#
Ê
Ê
'0" sec x dx
'0" sec x dx (1 0) "# Š 13 ‹ $
Ê
Thus a lower bound is 76 .
'ab f(x) dx is a constant K. Thus'ab av(f) dx œ 'ab K dx 'ab av(f) dx œ (b a)K œ (b a) † b " a 'ab f(x) dx œ 'ab f(x) dx.
73. Yes, for the following reasons: av(f) œ œ K(b a) Ê
" ba
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
'0" sec x dx 76 .
312
Chapter 5 Integration
74. All three rules hold. The reasons: On any interval [aß b] on which f and g are integrable, we have: (a) av(f g) œ
" ba
'ab [f(x) g(x)] dx œ b " a ”'ab f(x) dx 'ab g(x) dx• œ b " a 'ab f(x) dx b " a 'ab g(x) dx
œ av(f) av(g) (b) av(kf) œ (c) av(f) œ
" ba
" ba
'ab kf(x) dx œ b " a ”k 'ab f(x) dx• œ k ” b " a 'ab f(x) dx• œ k av(f)
'ab f(x) dx Ÿ b " a 'ab g(x) dx since f(x) Ÿ g(x) on [aß b], and b " a 'ab g(x) dx œ av(g).
Therefore, av(f) Ÿ av(g). ba n and let ck be the right n ab a b × and ck œ a kabn ab . n
75. Consider the partition P that subdivides the interval Òa, bÓ into n subintervals of width ˜x œ endpoint of each subinterval. So the partition is P œ Öa, a n
n
kœ"
kœ" b
We get the Riemann sum ! fack b˜x œ ! c † this expression remains cab ab. Thus,
ba n
œ
ba n , n
c ab a b ! " n kœ"
a
œ
#a b a b , n
c ab a b n
...,a
† n œ cab ab. As n Ä _ and mPm Ä !
'a c dx œ cab ab.
76. Consider the partition P that subdivides the interval Òa, bÓ into n subintervals of width ˜x œ endpoint of each subinterval. So the partition is P œ n
n
We get the Riemann sum ! fack b˜x œ ! ck# ˆ b n a ‰ œ
n ba ! # a n Œ kœ"
kœ" n #a a b a b ! k n kœ"
œ ab aba# aab ab# †
n" n
kœ" n ab a b # ! # k n# kœ"
ab a b $ '
†
œ
b a n
ba n
and let ck be the right
Öa, a b n a , a #abn ab , . . ., a nabn ab × and ck œ a kabn ab . n n # # # œ b n a ! Ša kabn ab ‹ œ bn a ! Ša# #akabn ab k abn# ab ‹ kœ" kœ" † na#
an "ba#n "b n#
#a a b a b# n#
†
n a n "b #
ab a b $ n$
†
nan "ba#n "b '
$ " " n" ab ab$ # n n# † " ' " ab a b $ †# ' b $ $ x# dx œ b$ a$ . a
œ ab aba# aab ab# †
As n Ä _ and mPm Ä ! this expression has value ab aba# aab ab# † " œ ba# a$ ab# #a# b a$ "$ ab$ $b# a $ba# a$ b œ
b$ $
a$ $.
Thus,
'
77. (a) U œ max" ?x max# ?x á maxn ?x where max" œ f(x" ), max# œ f(x# ), á , maxn œ f(xn ) since f is increasing on [aß b]; L œ min" ?x min# ?x á minn ?x where min" œ f(x! ), min# œ f(x" ), á , minn œ f(xnc1 ) since f is increasing on [aß b]. Therefore U L œ (max" min" ) ?x (max# min# ) ?x á (maxn minn ) ?x œ (f(x" ) f(x! )) ?x (f(x# ) f(x" ))?x á (f(xn ) f(xnc1 )) ?x œ (f(xn ) f(x! )) ?x œ (f(b) f(a)) ?x. (b) U œ max" ?x" max# ?x# á maxn ?xn where max" œ f(x" ), max# œ f(x# ), á , maxn œ f(xn ) since f is increasing on[aß b]; L œ min" ?x" min# ?x# á minn ?xn where min" œ f(x! ), min# œ f(x" ), á , minn œ f(xnc1 ) since f is increasing on [aß b]. Therefore U L œ (max" min" ) ?x" (max# min# ) ?x# á (maxn minn ) ?xn œ (f(x" ) f(x! )) ?x" (f(x# ) f(x" ))?x# á (f(xn ) f(xnc1 )) ?xn Ÿ (f(x" ) f(x! )) ?xmax (f(x# ) f(x" )) ?xmax á (f(xn ) f(xnc1 )) ?xmax . Then U L Ÿ (f(xn ) f(x! )) ?xmax œ (f(b) f(a)) ?xmax œ kf(b) f(a)k ?xmax since f(b) f(a). Thus lim (U L) œ lim (f(b) f(a)) ?xmax œ 0, since ?xmax œ lPl . lPl Ä 0
lPl Ä 0
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Section 5.3 The Definite Integral
313
78. (a) U œ max" ?x max# ?x á maxn ?x where max" œ f(x! ), max# œ f(x" ), á , maxn œ f(xnc" ) since f is decreasing on [aß b]; L œ min" ?x min# ?x á minn ?x where min" œ f(x" ), min# œ f(x# )ß á , minn œ f(xn ) since f is decreasing on [aß b]. Therefore U L œ (max" min" ) ?x (max# min# ) ?x á (maxn minn ) ?x œ (f(x! ) f(x" )) ?x (f(x" ) f(x# ))?x á (f(xn" ) f(xn )) ?x œ (f(x! ) f(xn )) ?x œ (f(a) f(b)) ?x. (b) U œ max" ?x" max# ?x# á maxn ?xn where max" œ f(x! ), max# œ f(x" ), á , maxn œ f(xn" ) since f is decreasing on[aß b]; L œ min" ?x" min# ?x# á minn ?xn where min" œ f(x" ), min# œ f(x# ), á , minn œ f(xn ) since f is decreasing on [aß b]. Therefore U L œ (max" min" ) ?x" (max# min# ) ?x# á (maxn minn ) ?xn œ (f(x! ) f(x" )) ?x" (f(x" ) f(x# ))?x# á (f(xn" ) f(xn )) ?xn Ÿ (f(x! ) f(xn )) ?xmax œ (f(a) f(b) ?xmax œ kf(b) f(a)k ?xmax since f(b) Ÿ f(a). Thus lim (U L) œ lim kf(b) f(a)k ?xmax œ 0, since ?xmax œ lPl . lPl Ä 0
lPl Ä 0
79. (a) Partition 0ß 1# ‘ into n subintervals, each of length ?x œ x# œ 2?x, á , xn œ n?x œ
1 #.
1 #n
with points x! œ 0, x" œ ?x,
Since sin x is increasing on 0ß 1# ‘ , the upper sum U is the sum of the areas
of the circumscribed rectangles of areas f(x" ) ?x œ (sin ?x)?x, f(x# ) ?x œ (sin 2?x) ?x, á , f(xn ) ?x œ (sin n?x) ?x. Then U œ (sin ?x sin 2?x á sin n?x) ?x œ ” œ”
1 cos ˆˆn " ‰ 1 ‰ cos 4n 1 # 2n 1 • ˆ #n ‰ # sin 4n
'!
1 cos ˆ 1 1 ‰‰ 1 ˆcos 4n # 4n 1 4n sin 4n
œ
1 cos ˆ 1 1 ‰ cos 4n # 4n sin 1
Š 14n ‹ 4n
1Î#
(b) The area is
œ
cos ?#x cosˆ ˆn #" ‰ ?x‰ • ?x # sin ?#x
sin x dx œ n lim Ä_
1 cos ˆ 1 1 ‰ cos 4n # 4n sin 1
Š 14n ‹
œ
1 cos 1# 1
œ 1.
4n
n
80. (a) The area of the shaded region is !˜xi † mi which is equal to L. iœ" n
(b) The area of the shaded region is !˜xi † Mi which is equal to U. iœ"
(c) The area of the shaded region is the difference in the areas of the shaded regions shown in the second part of the figure and the first part of the figure. Thus this area is U L. n
n
iœ"
iœ"
81. By Exercise 80, U L œ !˜xi † Mi !˜xi † mi where Mi œ maxÖfaxb on the ith subinterval× and n
n
iœ"
iœ"
mi œ minÖfaxb on the ith subinterval×. Thus U L œ !aMi mi b˜xi !% † ˜xi provided ˜xi $ for each n
n
iœ"
iœ"
i œ "ß Þ Þ Þ , n. Since !% † ˜xi œ % !˜xi œ %ab ab the result, U L %ab ab follows.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
314
Chapter 5 Integration
82. The car drove the first 150 miles in 5 hours and the second 150 miles in 3 hours, which means it drove 300 miles in 8 hours, for an average of 300 8 mi/hr œ 37.5 mi/hr. In terms of average values of functions, the function whose average value we seek is 30, 0 Ÿ t Ÿ 5 v(t) œ œ , and the average value is 50, 5 1 Ÿ 8 (30)(5) (50)(3) 8
œ 37.5.
83-88. Example CAS commands: Maple: with( plots ); with( Student[Calculus1] ); f := x -> 1-x; a := 0; b := 1; N :=[ 4, 10, 20, 50 ]; P := [seq( RiemannSum( f(x), x=a..b, partition=n, method=random, output=plot ), n=N )]: display( P, insequence=true ); 89-92. Example CAS commands: Maple: with( Student[Calculus1] ); f := x -> sin(x); a := 0; b := Pi; plot( f(x), x=a..b, title="#23(a) (Section 5.1)" ); N := [ 100, 200, 1000 ]; # (b) for n in N do Xlist := [ a+1.*(b-a)/n*i $ i=0..n ]; Ylist := map( f, Xlist ); end do: for n in N do # (c) Avg[n] := evalf(add(y,y=Ylist)/nops(Ylist)); end do; avg := FunctionAverage( f(x), x=a..b, output=value ); evalf( avg ); FunctionAverage(f(x),x=a..b,output=plot); # (d) fsolve( f(x)=avg, x=0.5 ); fsolve( f(x)=avg, x=2.5 ); fsolve( f(x)=Avg[1000], x=0.5 ); fsolve( f(x)=Avg[1000], x=2.5 ); 83-92. Example CAS commands: Mathematica: (assigned function and values for a, b, and n may vary) Sums of rectangles evaluated at left-hand endpoints can be represented and evaluated by this set of commands Clear[x, f, a, b, n]
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 5.4 The Fundamental Theorem of Calculus {a, b}={0, 1}; n =10; dx = (b a)/n; f = Sin[x]2 ; xvals =Table[N[x], {x, a, b dx, dx}]; yvals = f /.x Ä xvals; boxes = MapThread[Line[{{#1,0},{#1, #3},{#2, #3},{#2, 0}]&,{xvals, xvals dx, yvals}]; Plot[f, {x, a, b}, Epilog Ä boxes]; Sum[yvals[[i]] dx, {i, 1, Length[yvals]}]//N Sums of rectangles evaluated at right-hand endpoints can be represented and evaluated by this set of commands. Clear[x, f, a, b, n] {a, b}={0, 1}; n =10; dx = (b a)/n; f = Sin[x]2 ; xvals =Table[N[x], {x, a dx, b, dx}]; yvals = f /.x Ä xvals; boxes = MapThread[Line[{{#1,0},{#1, #3},{#2, #3},{#2, 0}]&,{xvals dx,xvals, yvals}]; Plot[f, {x, a, b}, Epilog Ä boxes]; Sum[yvals[[i]] dx, {i, 1,Length[yvals]}]//N Sums of rectangles evaluated at midpoints can be represented and evaluated by this set of commands. Clear[x, f, a, b, n] {a, b}={0, 1}; n =10; dx = (b a)/n; f = Sin[x]2 ; xvals =Table[N[x], {x, a dx/2, b dx/2, dx}]; yvals = f /.x Ä xvals; boxes = MapThread[Line[{{#1,0},{#1, #3},{#2, #3},{#2, 0}]&,{xvals dx/2, xvals dx/2, yvals}]; Plot[f, {x, a, b},Epilog Ä boxes]; Sum[yvals[[i]] dx, {i, 1, Length[yvals]}]//N 5.4 THE FUNDAMENTAL THEOREM OF CALCULUS 1.
'c (2x 5) dx œ cx# 5xd#! œ a0# 5(0)b a(2)# 5(2)b œ 6
2.
'c ˆ5 x# ‰ dx œ ’5x x4 “ %
0
2
4
3.
#
$
3
'
4
0
Š3x
x$ 4‹
'c ax$ 2x 3b dx œ ’ x4
%
'
6.
'
7.
'
1
# #
Š5(3)
4% 16 ‹
#
Š 3(0) #
(3)# 4 ‹
(0)% 16 ‹
œ
133 4
œ8 %
œ Š 24 2# 3(2)‹ Š (42) (2)# 3(2)‹ œ 12 %
"
$
ˆx# Èx‰ dx œ ’ x3 23 x$Î# “ œ ˆ "3 23 ‰ 0 œ 1 !
0
0
#
4# 4‹
œ Š 3(4) #
x# 3x“
2
5.
5
&
x$Î# dx œ 25 x&Î# ‘ ! œ
32
1
(5)&Î# 0 œ 2(5)$Î# œ 10È5
$#
'cc x2 2
2 5
x'Î& dx œ 5x"Î& ‘ " œ ˆ #5 ‰ (5) œ
1
8.
% x% 16 “ !
#
dx œ ’ 3x#
2
4.
œ Š5(4)
#
dx œ
5 #
'cc 2x# dx œ c2x" d " ˆ 2 ‰ ˆ 2 ‰ # œ 1 # œ 1 1
2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
315
316
Chapter 5 Integration
'
1
9.
'
1
10.
'
1Î3
11.
'
51Î6
12.
'
31Î4
13.
'
1Î3
14.
15.
'
0
0
sin x dx œ [cos x]1! œ (cos 1) (cos 0) œ (1) (1) œ 2 (1 cos x) dx œ [x sin x]1! œ (1 sin 1) (0 sin 0) œ 1
0
1Î$
1Î6
1Î4
0
œ ˆ2 tan ˆ 13 ‰‰ (2 tan 0) œ 2È3 0 œ 2È3
2 sec# x dx œ [2 tan x]!
&1Î' csc# x dx œ [cot x]1Î' œ ˆcot ˆ 561 ‰‰ ˆcot ˆ 16 ‰‰ œ ŠÈ3‹ ŠÈ3‹ œ 2È3
$1Î%
csc ) cot ) d) œ [csc )]1Î% œ ˆcsc ˆ 341 ‰‰ ˆcsc ˆ 14 ‰‰ œ È2 ŠÈ2‹ œ 0 1Î$
4 sec u tan u du œ [4 sec u]!
0
" cos 2t # 1Î2
dt œ
'
0
ˆ"
1Î2 #
" #
œ 4 sec ˆ 13 ‰ 4 sec 0 œ 4(2) 4(1) œ 4
cos 2t‰ dt œ "# t
" 4
!
sin 2t‘ 1Î# œ ˆ "# (0)
" 4
sin 2(0)‰ ˆ "# ˆ 1# ‰
" 4
sin 2 ˆ 1# ‰‰
œ 14 Î 1Î$ 2t 'c ÎÎ " cos dt œ ' ˆ "# "# cos 2t‰ dt œ "# t 4" sin 2t‘ 1Î$ # Î 1 3
16.
1 3
1 3
1 3
œ ˆ "# ˆ 13 ‰
" 4
sin 2 ˆ 13 ‰‰ ˆ #" ˆ 13 ‰
" 4
sin 2 ˆ 13 ‰‰ œ
1 6
" 4
sin 231
17.
'c
18.
'cÎ Î% ˆ4 sec# t t1 ‰ dt œ 'Î Î% a4 sec# t 1t# b dt œ 4 tan t 1t ‘ 11Î%Î$
1Î#
1Î#
$
a8y# sin yb dy œ ’ 8y3 cos y“
1Î#
1Î#
1
œŒ
8 ˆ 1# ‰ 3
$
8 ˆ 1# ‰ 3
cos 1# Œ
1 6 $
œ Š4 tan ˆ
cos ˆ 1# ‰ œ
1‰ 4
Š4 tan ˆ 13 ‰
1 ˆ 13 ‰ ‹
È3 4
21 $ 3
œ (4(1) 4) Š4 ŠÈ3‹ 3‹ œ 4È3 3
'"" (r 1)# dr œ '"" ar# 2r 1b dr œ ’ r3 r# r“ " œ Š (31)
20.
'È (t 1) at# 4b dt œ 'È at$ t# 4t 4b dt œ ’ t4 t3 2t# 4t“ÈÈ$
$
"
È3
È3
3
œ
%
$
$
(1)# (1)‹ Š 13 1# 1‹ œ 38
$
3
%
ŠÈ3‹
$
ŠÈ3‹
4
3
21.
'È" Š u#
" u& ‹
22.
' " ˆ v"
"‰ v%
23.
'
(
2
1
1 3
1 3
1 ˆ 14 ‰ ‹
19.
È2
sin ˆ 321 ‰ œ
1
#
1 3
1Î2
" 4
$
s# È s s#
(
2
dv œ
ds œ
'
1
È 3 ‹
#
Š 2 ŠÈ3‹ 4È3
" du œ 'È Š u#
%
&
u ‹ du œ
u) ’ 16
4
$
" " 4u% “È#
ŠÈ3‹
È2
ˆ1 s$Î# ‰ ds œ ’s
#
)
œ
1) Š 16
$
È# 2 “ Ès "
#
2 ŠÈ3‹ 4 ŠÈ3‹ œ 10È3
3
" ' " av$ v% b dv œ 2v1 3v" ‘ ""Î# œ Š 2(1) 1Î2
$
#
œ È 2
" 4(1)% ‹
" 3(1)$ ‹
2 É È2
ŠÈ2‹ 16
" % 4 ŠÈ2‹
" $ 3 ˆ "# ‰
œ 34
Œ
" # 2 ˆ "# ‰
Š1
2 È1 ‹
œ È2 2$Î% 1
œ È2 %È8 1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
œ 56
Section 5.4 The Fundamental Theorem of Calculus 24.
'
4
1 Èu Èu 9
du œ
'
4
9
ˆu"Î# 1‰ du œ 2Èu u‘ % œ Š2È4 4‹ Š2È9 9‹ œ 3 *
'c% kxk dx œ '%! kxk dx '! 4
25.
4
kxk dx œ
'%! x dx '!
4
#
x dx œ ’ x# “
œ 16 26.
'
1
" ! #
acos x kcos xk b dx œ 1 #
œ sin
27. (a)
!
d dx
28. (a)
'
(b)
d dx
29. (a)
'
(b)
d dt
30. (a)
'
!
" # (cos
x cos x) dx
'
1
" 1Î# #
#
’ x# “ œ Š 0# !
(cos x cos x) dx œ
'
1Î#
!
(4)# # ‹
#
Š 4#
Èx
cos t dt œ [sin t]! œ sin Èx sin 0 œ sin Èx Ê
Èx
'
Œ
sin x
1
!
d dx
Œ
'
Èx
!
cos t dt œ
sin x
'
!
d ˆÈ ‰‰ cos t dt œ ˆcos Èx‰ ˆ dx x œ ˆcos Èx ‰ ˆ "# x"Î# ‰ œ
3t# dt œ ct$ d "
Πt%
1Î#
d dx
ˆsin Èx‰ œ cos Èx ˆ "# x"Î# ‰
sin x
Èu du œ t%
!
tan )
!
d dx
Œ
'
sin x
1
3t# dt œ
d dx
asin$ x 1b œ 3 sin# x cos x
d 3t# dt œ a3 sin# xb ˆ dx (sin x)‰ œ 3 sin# x cos x
1
Œ'
œ sin$ x 1 Ê
cos Èx 2È x
'
t%
!
t%
u"Î# du œ 23 u$Î# ‘ ! œ
2 3
at% b
$Î#
0œ
2 ' 3 t
Ê
d dt
Œ'
Œ
'
t%
!
Èu du œ
d dt
ˆ 23 t' ‰ œ 4t&
Èu du œ Èt% ˆ dtd at% b‰ œ t# a4t$ b œ 4t&
) sec# y dy œ [tan y]tan œ tan (tan )) 0 œ tan (tan )) Ê !
d d)
tan )
!
sec# y dy œ
d d)
(tan (tan )))
œ asec# (tan ))b sec# ) (b)
d d)
'
Œ
tan )
!
sec# y dy œ asec# (tan ))b ˆ dd) (tan ))‰ œ asec# (tan ))b sec# )
31. y œ
'
33. y œ
'È sin t# dt œ '
34. y œ
'
35. y œ
'
36. y œ
'
x
!
È1 t# dt Ê
Èx
!
x
x#
cos Èt dt Ê
sin x
dt È1 t#
!
!
tan x
dt 1 t#
, kxk Ê
dy dx
œ È1 x#
dy dx
!
!
0# #‹
cos x dx œ [sin x]!
cos Èx 2È x
œ (b)
1Î#
%
#
%
sin 0 œ 1
Èx
'
'
!
dy dx
1 #
sin t# dt Ê
32. y œ dy dx
'
1
x
" t
dt Ê
dy dx
œ
" x
,x0
#
d ˆÈ ‰‰ œ Šsin ˆÈx‰ ‹ ˆ dx x œ (sin x) ˆ "# x"Î# ‰ œ 2sinÈxx
d œ Šcos Èx# ‹ ˆ dx ax# b‰ œ 2x cos kxk
Ê
dy dx
œ
" È1 sin# x
d ˆ dx (sin x)‰ œ
" Ècos# x
(cos x) œ
cos x kcos xk
œ
cos x cos x
œ 1 since kxk
" d ‰ ˆ dx œ ˆ 1 tan (tan x)‰ œ ˆ sec"# x ‰ asec# xb œ 1 #x
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
1 #
317
318
Chapter 5 Integration
37. x# 2x œ 0 Ê x(x 2) œ 0 Ê x œ 0 or x œ 2; Area œ
'$# ax# 2xbdx '#! ax# 2xbdx '!# ax# 2xbdx $
œ ’ x3 x# “ œ ŠŠ
(2)$ 3
# $
$
’ x3 x# “ #
(2) ‹ Š
!
$
#
(3)$ 3
’ x3 x# “
#
!
#
(3) ‹‹
$
ŠŠ 03 0# ‹ Š (32) (2)# ‹‹ $
$
$
ŠŠ 23 2# ‹ Š 03 0# ‹‹ œ
28 3
38. 3x# 3 œ 0 Ê x# œ 1 Ê x œ „ 1; because of symmetry about the y-axis, Area œ 2 Œ
'!" a3x# 3bdx '"# a3x# 3bdx
"
#
2 Š cx$ 3xd ! cx$ 3xd " ‹ œ 2 c aa1$ 3(1)b a0$ 3(0)bb aa2$ 3(2)b a1$ 3(1)bd œ 2(6) œ 12
39. x$ 3x# 2x œ 0 Ê x ax# 3x 2b œ 0 Ê x(x 2)(x 1) œ 0 Ê x œ 0, 1, or 2; Area œ
'!" ax$ 3x# 2xbdx '"# ax$ 3x# 2xbdx "
%
%
œ ’ x4 x$ x# “ ’ x4 x$ x# “ !
%
# "
%
œ Š 14 1$ 1# ‹ Š 04 0$ 0# ‹ %
%
’Š 24 2$ 2# ‹ Š 14 1$ 1# ‹“ œ
" #
40. x$ 4x œ 0 Ê x ax# 4b œ 0 Ê x(x 2)(x 2) œ 0 Ê x œ 0, 2, or 2. Area œ œ
% ’ x4
#
2x “ %
! #
% ’ x4
'c! ax$ 4xbdx '!# ax$ 4xbdx 2
#
#
%
2x “ œ Š 04 2(0)# ‹ !
Š (42) 2(2)# ‹ ’Š 24 2(2)# ‹ Š 04 2(0)# ‹“ œ 8 %
41. x"Î$ œ 0 Ê x œ 0; Area œ
%
'c! x"Î$ dx '!) x"Î$ dx "
! ) œ 34 x%Î$ ‘ " 34 x%Î$ ‘ ! œ ˆ 34 (0)%Î$ ‰ ˆ 34 (1)%Î$ ‰ ˆ 34 (8)%Î$ ‰ ˆ 34 (0)%Î$ ‰
œ
51 4
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 5.4 The Fundamental Theorem of Calculus 42. x"Î$ x œ 0 Ê x"Î$ ˆ1 x#Î$ ‰ œ 0 Ê x"Î$ œ 0 or 1 x#Î$ œ 0 Ê x œ 0 or 1 œ x#Î$ Ê x œ 0 or 1 œ x# Ê x œ 0 or „ 1; Area œ œ
'c! ˆx"Î$ x‰dx '!" ˆx"Î$ x‰dx '") ˆx"Î$ x‰dx "
%Î$
’ 34
x
! x# # “ "
œ ’Š 34 (0)%Î$
’ 34 x%Î$
0# #‹
" x# # “!
’ 43 x%Î$ (1)# # ‹“
Š 34 (1)%Î$
’Š 34 (1)%Î$
1# #‹
Š 34 (0)%Î$
0# # ‹“
’Š 34 (8)%Î$
8# #‹
Š 34 (1)%Î$
1# # ‹“
œ
" 4
" 4
ˆ2!
$ 4
#" ‰ œ
) x# # “"
83 4
43. The area of the rectangle bounded by the lines y œ 2, y œ 0, x œ 1, and x œ 0 is 21. The area under the curve y œ 1 cos x on [0ß 1] is
'!
1
(1 cos x) dx œ [x sin x]!1 œ (1 sin 1) (0 sin 0) œ 1. Therefore the area of
the shaded region is 21 1 œ 1. 44. The area of the rectangle bounded by the lines x œ 16 , x œ " #
51 6 ,
y œ sin
ˆ 561 16 ‰ œ 13 . The area under the curve y œ sin x on 16 ß 561 ‘ is
œ ˆcos
51 ‰ 6
È3 # ‹
ˆcos 16 ‰ œ Š
È3 #
'
1 6
œ
51Î6
1Î6
" #
œ sin
51 6
, and y œ 0 is &1Î'
sin x dx œ [cos x]1Î'
œ È3. Therefore the area of the shaded region is È3 13 .
45. On 14 ß 0‘ : The area of the rectangle bounded by the lines y œ È2, y œ 0, ) œ 0, and ) œ 14 is È2 ˆ 14 ‰ œ
1È2 4
. The area between the curve y œ sec ) tan ) and y œ 0 is
'c
!
1Î4
sec ) tan ) d) œ [sec )]!1Î%
œ (sec 0) ˆsec ˆ 14 ‰‰ œ È2 1. Therefore the area of the shaded region on 14 ß !‘ is
1È2 4
On 0ß 14 ‘ : The area of the rectangle bounded by ) œ 14 , ) œ 0, y œ È2, and y œ 0 is È2 ˆ 14 ‰ œ under the curve y œ sec ) tan ) is of the shaded region on !ß 14 ‘ is È
'
1Î4
!
1È2 4
1Î%
sec ) tan ) d) œ [sec )]!
œ sec
1 4
ŠÈ2 1‹ .
1È2 4
. The area
sec 0 œ È2 1. Therefore the area
ŠÈ2 1‹ . Thus, the area of the total shaded region is
È
1È2 #
Š 1 4 2 È2 1‹ Š 1 4 2 È2 1‹ œ
.
46. The area of the rectangle bounded by the lines y œ 2, y œ 0, t œ 14 , and t œ 1 is 2 ˆ1 ˆ 14 ‰‰ œ 2 area under the curve y œ sec# t on 14 ß !‘ is under the curve y œ 1 t# on [!ß "] is
'c
!
1Î4
"
$
œ
dt 3 œ 0 3 œ 3 Ê (d) is a solution to this problem.
dy dx
œ
48. y œ
'c sec t dt 4
Ê
dy dx
œ sec x and y(1) œ
49. y œ
'
sec t dt 4 Ê
dy dx
œ sec x and y(0) œ
x
1
!
x
and y(1) œ
'
" t
dt 3 Ê
" x
1
1
Thus, the total
. Therefore the area of the shaded region is ˆ2 1# ‰
'
" 1 t
$
5 3
47. y œ
x
$
!
2 3
'
!
. The
! ˆ 1‰ sec# t dt œ [tan t] 1Î% œ tan 0 tan 4 œ 1. The area
'! a1 t# b dt œ ’t t3 “ " œ Š1 13 ‹ Š0 03 ‹ œ 32 .
area under the curves on 14 ß "‘ is 1
1 #
'cc sec t dt 4 œ 0 4 œ 4 1
1
!
5 3
œ
" 3
1 #
.
Ê (c) is a solution to this problem.
sec t dt 4 œ 0 4 œ 4 Ê (b) is a solution to this problem.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
319
320
Chapter 5 Integration
50. y œ
'
51. y œ
'
53. s œ
'
x
" " t
"
" t
dt 3 œ 0 3 œ 3 Ê (a) is a solution to this problem.
'
'c ÎÎ
b 2 b 2
$
ˆ bh #
bh ‰ 6
Š2
b Œh ˆ # ‰
2 (x 1)# ‹
bh ‰ 6
dx œ 2
'
$
!
œ
bh 3
t
t!
È1 t# dt 2 g(x) dx v!
$
2 3
bh
" (x 1)# ‹
Š1
"
bÎ2
4h ˆ #b ‰ 3b#
œ bh
x
4hx$ 3b# “ bÎ2
ˆh ˆ 4h ‰ # ‰ dx œ ’hx b# x
œ ˆ bh #
!
"
54. v œ
4h ˆ b# ‰ 3b#
$
'
f(x) dx s!
œ Œhˆ #b ‰
'
and y(1) œ
'
55. Area œ
56. r œ
" x
52. y œ
#
t!
œ
dy dx
sec t dt 3
x
t
dt 3 Ê
$
dx œ 2 x ˆ x11 ‰‘ ! œ 2 ’Š3
" (3 1) ‹
Š0
" (0 1) ‹“
œ 2 3 "4 1‘ œ 2 ˆ2 4" ‰ œ 4.5 or $4500 57.
dc dx
œ
" #È x
œ
" #
x"Î# Ê c œ
'
x
!
" "Î# dt # t
œ t"Î# ‘ 0 œ Èx x
c(100) c(1) œ È100 È1 œ $9.00 58. By Exercise 57, c(400) c(100) œ È400 È100 œ 20 10 œ $10.00 59. (a) v œ (b) a œ (c) s œ (d) (e) (f) (g)
ds dt df dt
'
!
œ
d dt
'
t
!
f(x) dx œ f(t) Ê v(5) œ f(5) œ 2 m/sec
is negative since the slope of the tangent line at t œ 5 is negative 3
f(x) dx œ
" #
(3)(3) œ
9 #
m since the integral is the area of the triangle formed by y œ f(x), the x-axis,
and x œ 3 t œ 6 since from t œ 6 to t œ 9, the region lies below the x-axis At t œ 4 and t œ 7, since there are horizontal tangents there Toward the origin between t œ 6 and t œ 9 since the velocity is negative on this interval. Away from the origin between t œ 0 and t œ 6 since the velocity is positive there. Right or positive side, because the integral of f from 0 to 9 is positive, there being more area above the x-axis than below it.
60. (a) v œ (b) a œ
dg dt df dt
œ
d dt
'
!
t
g(x) dx œ g(t) Ê v(3) œ g(3) œ 0 m/sec.
is positive, since the slope of the tangent line at t œ 3 is positive
(c) At t œ 3, the particle's position is
'
!
$
g(x) dx œ
" #
(3)(6) œ 9
(d) The particle passes through the origin at t œ 6 because s(6) œ
'
!
'
g(x) dx œ 0
(e) At t œ 7, since there is a horizontal tangent there (f) The particle starts at the origin and moves away to the left for 0 t 3. It moves back toward the origin for 3 t 6, passes through the origin at t œ 6, and moves away to the right for t 6. (g) Right side, since its position at t œ 9 is positive, there being more area above the x-axis than below it at t œ *.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 5.4 The Fundamental Theorem of Calculus 61. k 0 Ê one arch of y œ sin kx will occur over the interval 0ß 1k ‘ Ê the area œ œ "k cos ˆk ˆ 1k ‰‰ ˆ k" cos (0)‰ œ 62. lim x"$ xÄ!
63.
'
64.
'
x
1
x
!
'
!
x
t%
t# dt "
œ lim
' x %t# dt ! t " x$
xÄ!
x
66.
1
x#
" k
x#
d dx
xÄ!
'
d dx
'
!
1
x
x
f(t) dt œ
d dx
ax# 2x 1b œ 2x 2
f(t) dt œ cos 1x 1x sin 1x Ê f(4) œ cos 1(4) 1(4) sin 1(4) œ 1
Ê f w (x) œ 1 (x9 1) œ
9 x 2
Ê f w (1) œ 3; f(1) œ 2
'#"" 1 9 t dt œ 2 0 œ 2;
sec (t 1) dt Ê gw (x) œ asec ax# 1bb (2x) œ 2x sec ax# 1b Ê gw (1) œ 2(1) sec a(1)# 1b #
a"b " œ 2; g(1) œ 3 ' sec (t 1) dt œ 3 ' sec (t 1) dt œ 3 0 œ 3; L(x) œ 2(x (1)) g(1) 1
1
œ 2(x 1) 3 œ 2x 1 67. (a) (b) (c) (d) (e) (f) (g)
True: since f is continuous, g is differentiable by Part 1 of the Fundamental Theorem of Calculus. True: g is continuous because it is differentiable. True, since gw (1) œ f(1) œ 0. False, since gww (1) œ f w (1) 0. True, since gw (1) œ 0 and gww (1) œ f w (1) 0. False: gww (x) œ f w (x) 0, so gww never changes sign. True, since gw (1) œ f(1) œ 0 and gw (x) œ f(x) is an increasing function of x (because f w (x) 0).
68. (a) True: by Part 1 of the Fundamental Theorem of Calculus, hw (x) œ f(x). Since f is differentiable for all x, h has a second derivative for all x. (b) True: they are continuous because they are differentiable. (c) True, since hw (1) œ f(1) œ 0. (d) True, since hw (1) œ 0 and hww (1) œ f w (1) 0. (e) False, since hww (1) œ f w (1) 0. (f) False, since hww (x) œ f w (x) 0 never changes sign. (g) True, since hw (1) œ f(1) œ 0 and hw (x) œ f(x) is a decreasing function of x (because f w (x) 0). 69.
1 Îk
cos kx‘ !
%
L(x) œ 3(x 1) f(1) œ 3(x 1) 2 œ 3x 5 g(x) œ 3 '
sin kx dx œ
2 k
xÄ!
f(t) dt œ x cos 1x Ê f(x) œ
'# " 1 9 t dt
!
1Îk
œ lim x$x#" œ lim $ax%" "b œ _.
f(t) dt œ x# 2x 1 Ê f(x) œ
65. f(x) œ 2
'
321
70. The limit is 3x#
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322
Chapter 5 Integration
71-74. Example CAS commands: Maple: with( plots ); f := x -> x^3-4*x^2+3*x; a := 0; b := 4; F := unapply( int(f(t),t=a..x), x ); # (a) p1 := plot( [f(x),F(x)], x=a..b, legend=["y = f(x)","y = F(x)"], title="#71(a) (Section 5.4)" ): p1; dF := D(F); # (b) q1 := solve( dF(x)=0, x ); pts1 := [ seq( [x,f(x)], x=remove(has,evalf([q1]),I) ) ]; p2 := plot( pts1, style=point, color=blue, symbolsize=18, symbol=diamond, legend="(x,f(x)) where F '(x)=0" ): display( [p1,p2], title="71(b) (Section 5.4)" ); incr := solve( dF(x)>0, x ); # (c) decr := solve( dF(x) x^2; f := x -> sqrt(1-x^2); F := unapply( int( f(t), t=a..u(x) ), x ); dF := D(F); # (b) cp := solve( dF(x)=0, x ); solve( dF(x)>0, x ); solve( dF(x) x^3/3-x^2/2-2*x+1/3; g := x -> x-1; plot( [f(x),g(x)], x=-5..5, legend=["y = f(x)","y = g(x)"], title="#91(a) (Section 5.6)" ); q1 := [ -5, -2, 1, 4 ]; # (b) q2 := [seq( fsolve( f(x)=g(x), x=q1[i]..q1[i+1] ), i=1..nops(q1)-1 )]; for i from 1 to nops(q2)-1 do # (c) area[i] := int( abs(f(x)-g(x)),x=q2[i]..q2[i+1] ); end do; add( area[i], i=1..nops(q2)-1 ); # (d) Mathematica: (assigned functions may vary) Clear[x, f, g] f[x_] = x2 Cos[x] g[x_] = x3 x Plot[{f[x], g[x]}, {x, 2, 2}] After examining the plots, the initial guesses for FindRoot can be determined. pts = x/.Map[FindRoot[f[x]==g[x],{x, #}]&, {1, 0, 1}] i1=NIntegrate[f[x] g[x], {x, pts[[1]], pts[[2]]}] i2=NIntegrate[f[x] g[x], {x, pts[[2]], pts[[3]]}] i1 i2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
343
344
Chapter 5 Integration
CHAPTER 5 PRACTICE EXERCISES 1. (a) Each time subinterval is of length ?t œ 0.4 sec. The distance traveled over each subinterval, using the midpoint rule, is ?h œ "# avi vib1 b ?t, where vi is the velocity at the left endpoint and vib1 the velocity at
the right endpoint of the subinterval. We then add ?h to the height attained so far at the left endpoint vi to arrive at the height associated with velocity vib1 at the right endpoint. Using this methodology we build the following table based on the figure in the text: t (sec) 0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0 4.4 4.8 5.2 5.6 6.0 v (fps) 0 10 25 55 100 190 180 165 150 140 130 115 105 90 76 65 h (ft) 0 2 9 25 56 114 188 257 320 378 432 481 525 564 592 620.2 t (sec) v (fps) h (ft)
6.4 50 643.2
6.8 37 660.6
7.2 25 672
7.6 12 679.4
8.0 0 681.8
NOTE: Your table values may vary slightly from ours depending on the v-values you read from the graph. Remember that some shifting of the graph occurs in the printing process. The total height attained is about 680 ft. (b) The graph is based on the table in part (a).
2. (a) Each time subinterval is of length ?t œ 1 sec. The distance traveled over each subinterval, using the midpoint rule, is ?s œ "# avi vib1 b ?t, where vi is the velocity at the left, and vib1 the velocity at the
right, endpoint of the subinterval. We then add ?s to the distance attained so far at the left endpoint vi to arrive at the distance associated with velocity vib1 at the right endpoint. Using this methodology we build the table given below based on the figure in the text, obtaining approximately 26 m for the total distance traveled: t (sec) 0 1 2 3 4 5 6 7 8 9 10 v (m/sec) 0 0.5 1.2 2 3.4 4.5 4.8 4.5 3.5 2 0 s (m) 0 0.25 1.1 2.7 5.4 9.35 14 18.65 22.65 25.4 26.4
(b) The graph shows the distance traveled by the moving body as a function of time for 0 Ÿ t Ÿ 10.
3. (a) (c)
10
!
kœ1 10
ak 4
œ
" 4
10
! ak œ
kœ1
" 4
(2) œ #"
(b)
10
10
10
kœ1
kœ1
kœ1
10
10
10
kœ1
kœ1
kœ1
! (bk 3ak ) œ ! bk 3 ! ak œ 25 3(2) œ 31
! (ak bk 1) œ ! ak ! bk ! " œ 2 25 (1)(10) œ 13
kœ1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 5 Practice Exercises 10
10
kœ1
kœ1
! ˆ 5 bk ‰ œ ! #
(d)
20
5 #
10
! bk œ kœ1
20
kœ1 20
(b)
kœ1
! ˆ" #
(c)
kœ1 20
2bk ‰ 7
20
œ !
kœ1 20
" #
20
! bk œ
2 7
kœ1 20
" #
20
20
20
kœ1
kœ1
kœ1
! (ak bk ) œ ! ak ! bk œ 0 7 œ 7
(20) 27 (7) œ 8
! aak 2b œ ! ak ! 2 œ 0 2(20) œ 40
(d)
kœ1
kœ1
kœ1 " #
5. Let u œ 2x 1 Ê du œ 2 dx Ê 5
1
'
(2x 1)"Î# dx œ
9
1
3
1
x ax# 1b
7. Let u œ
"Î$
'
dx œ
8
0
du œ dx; x œ 1 Ê u œ 1, x œ 5 Ê u œ 9 *
u"Î# ˆ "# du‰ œ u"Î# ‘ " œ 3 1 œ 2
6. Let u œ x# 1 Ê du œ 2x dx Ê
'
(10) 25 œ 0
! 3ak œ 3 ! ak œ 3(0) œ 0
4. (a)
'
5 #
" #
du œ x dx; x œ 1 Ê u œ 0, x œ 3 Ê u œ 8 )
u"Î$ ˆ "# du‰ œ 38 u%Î$ ‘ ! œ
3 8
(16 0) œ 6
Ê 2 du œ dx; x œ 1 Ê u œ 1# , x œ 0 Ê u œ 0
x 2
' cos ˆ x# ‰ dx œ ' Î (cos u)(2 du) œ [2 sin u]!1Î# œ 2 sin 0 2 sin ˆ 1# ‰ œ 2(0 (1)) œ 2 0
0
1
1 2
8. Let u œ sin x Ê du œ cos x dx; x œ 0 Ê u œ 0, x œ
'
1Î2
0
(sin x)(cos x) dx œ
(e)
10. (a) (c) (e)
#
u du œ ’ u2 “ œ !
Ê uœ1
" #
2
2
5
2
2
(c)
0
"
'c f(x) dx œ "3 'c 3 f(x) dx œ 3" (12) œ 4 (b) ' f(x) dx œ ' f(x) dx ' f(x) dx œ 6 4 œ 2 c c c ' g(x) dx œ 'c g(x) dx œ 2 (d) ' (1 g(x)) dx œ 1 ' g(x) dx œ 1(2) œ 21 c c 'c Š f(x) 5 g(x) ‹ dx œ 5" 'c f(x) dx 5" 'c g(x) dx œ 5" (6) 5" (2) œ 85 2
9. (a)
'
1
1 #
' ' '
2
5
5
2
5
5
5
2
2
2
2
0 0
2
' 7 g(x) dx œ "7 (7) œ 1 f(x) dx œ ' f(x) dx œ 1 [g(x) 3 f(x)] dx œ ' g(x) dx 3' g(x) dx œ
2
" 7
(b)
0 2
(d)
0
2
0
5
2
0
0
2
' '
2
2
2
5
5
2
2
2
1 2
0
g(x) dx œ
'
0
2
g(x) dx
È2 f(x) dx œ È2
'
0
2
'
0
1
g(x) dx œ 1 2 œ 1
f(x) dx œ È2 (1) œ 1È2
f(x) dx œ 1 31
11. x# 4x 3 œ 0 Ê (x 3)(x 1) œ 0 Ê x œ 3 or x œ 1; Area œ
'
0
1
ax# 4x 3b dx "
$
'
1
$
3
ax# 4x 3b dx
œ ’ x3 2x# 3x“ ’ x3 2x# 3x“ !
$ "
$
œ ’Š "3 2(1)# 3(1)‹ 0“ $
$
’Š 33 2(3)# 3(3)‹ Š 13 2(1)# 3(1)‹“ œ ˆ "3 1‰ 0 ˆ 3" 1‰‘ œ
8 3
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
345
346
Chapter 5 Integration
12. 1
x# 4
œ 0 Ê 4 x# 0 Ê x œ „ 2;
Area œ
'c Š1 x4 ‹ dx ' 2
2
œ ’x
2
# x$ 12 “ # 2$ 12 ‹
œ ’Š2
3
#
’x
Š1
x# 4‹
dx
$ x$ 12 “ #
Š2
(2)$ 12 ‹“
œ 43 ˆ 43 ‰‘ ˆ 34 43 ‰ œ
’Š3
3$ 12 ‹
2$ 12 ‹“
Š2
13 4
13. 5 5x#Î$ œ 0 Ê 1 x#Î$ œ 0 Ê x œ „ 1; Area œ
'c ˆ5 5x#Î$ ‰ dx ' 1
1
1
8
ˆ5 5x#Î$ ‰ dx
" ) œ 5x 3x&Î$ ‘ " 5x 3x&Î$ ‘ " œ ˆ5(1) 3(1)&Î$ ‰ ˆ5(1) 3(1)&Î$ ‰‘
ˆ5(8) 3(8)&Î$ ‰ ˆ5(1) 3(1)&Î$ ‰‘ œ [2 (2)] [(40 96) 2] œ 62 14. 1 Èx œ 0 Ê x œ 1; Area œ œ œ œ
'
0
1
ˆ1 Èx‰ dx
4
1
ˆ1 Èx‰ dx
x 23 x$Î# ‘ " x 23 x$Î# ‘ % ! " ˆ1 23 (1)$Î# ‰ 0‘ ˆ4 23 " ˆ4 16 ‰ "‘ 3 3 3 œ 2
15. f(x) œ x, g(x) œ œ
'
'
2
1
ˆx
"‰ x#
œ
'
œ
Š 42
1
Šx
a œ 1, b œ 2 Ê A œ
'
b
[f(x) g(x)] dx
a
#
#
dx œ ’ x# x" “ œ ˆ 4# "# ‰ ˆ "# 1‰ œ 1
16. f(x) œ x, g(x) œ 2
" x# ,
(4)$Î# ‰ ˆ1 23 (1)$Î# ‰‘
"
" Èx
" Èx ‹dx
, a œ 1, b œ 2 Ê A œ # œ ’ x# 2Èx“
2È2‹ ˆ "# 2‰ œ
'
a
b
[f(x) g(x)] dx
#
" 7 4 È 2 #
'
# 17. f(x) œ ˆ1 Èx‰ , g(x) œ 0, a œ 0, b œ 1 Ê A œ
œ
'
0
1
ˆ1 2x"Î# x‰ dx œ ’x 43 x$Î# #
" x# # “!
x% #
" x( 7 “!
œ1
" #
" 7
œ
a
œ1
18. f(x) œ a1 x$ b , g(x) œ 0, a œ 0, b œ 1 Ê A œ œ ’x
b
'
a
b
'
[f(x) g(x)] dx œ 4 3
" #
œ
" 6
1
0
ˆ1 Èx‰# dx œ
(6 8 3) œ
[f(x) g(x)] dx œ
'
0
1
'
0
1
ˆ1 2Èx x‰ dx
" 6
#
a1 x$ b dx œ
'
0
9 14
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
1
a1 2x$ x' b dx
Chapter 5 Practice Exercises 19. f(y) œ 2y# , g(y) œ 0, c œ 0, d œ 3 Ê Aœ œ2
'
3
'
d
c
y# dy œ
0
'
[f(y) g(y)] dy œ 2 3
3
0
a2y# 0b dy
$
cy$ d ! œ 18
20. f(y) œ 4 y# , g(y) œ 0, c œ 2, d œ 2 Ê Aœ
'
d
c
# y$ 3 “ #
œ ’4y
'c a4 y# b dy 2
[f(y) g(y)] dy œ œ 2 ˆ8
8‰ 3
2
œ
32 3
y# 4
21. Let us find the intersection points:
y 2 4
œ
Ê y# y 2 œ 0 Ê (y 2)(y 1) œ 0 Ê y œ 1 or y œ 2 Ê c œ 1, d œ 2; f(y) œ Ê Aœ
'
d
c
y 2 4
2
#
1
œ
" 4
'c ay 2 y# b dy œ 4" ’ y#
œ
" 4
ˆ 4# 4 83 ‰ ˆ "# 2 3" ‰‘ œ
#
1
y# 4
'c Š y4 2 y4 ‹ dy
[f(y) g(y)] dy œ
2
, g(y) œ
2y 9 8
y# 4 4
22. Let us find the intersection points:
# y$ 3 “ "
œ
y 16 4
Ê y# y 20 œ 0 Ê (y 5)(y 4) œ 0 Ê y œ 4 or y œ 5 Ê c œ 4, d œ 5; f(y) œ Ê Aœ
'
d
c
[f(y) g(y)] dy œ
y 16 4
, g(y) œ
y# 4 4
'c Š y 416 y 4 4 ‹ dy 5
#
4
œ
" 4
'c ay 20 y# b dy œ "4 ’ y#
œ
" 4 " 4
125 ‰ ˆ 25 ‰‘ ˆ "#6 80 64 # 100 3 3 9 " 9 " ˆ # 180 63‰ œ 4 ˆ # 117‰ œ 8 (9 234) œ
œ
5
#
20y
4
23. f(x) œ x, g(x) œ sin x, a œ 0, b œ Ê Aœ
'
b
a
#
[f(x) g(x)] dx œ
œ ’ x# cos x“
1Î% !
#
œ Š 31#
'
1Î4
(x sin x) dx
1
24. f(x) œ 1, g(x) œ ksin xk , a œ 1# , b œ Ê Aœ œ
'c
œ2
0
'
a
b
[f(x) g(x)] dx œ
(1 sin x) dx
1Î2 1Î2
'
0
'
0
1Î2
243 8
1 4
0
È2 # ‹
& y$ 3 “ %
'c
1Î2
1Î2
1 2
a1 ksin xkb dx
(1 sin x) dx 1Î#
(1 sin x) dx œ 2[x cos x]!
œ 2 ˆ 1# 1‰ œ 1 2 Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
347
348
Chapter 5 Integration
25. a œ 0, b œ 1, f(x) g(x) œ 2 sin x sin 2x
'
Ê Aœ
1
0
(2 sin x sin 2x) dx œ 2 cos x
cos 2x ‘ 1 # !
œ 2 † (1) "# ‘ ˆ2 † 1 "# ‰ œ 4
26. a œ 13 , b œ 13 , f(x) g(x) œ 8 cos x sec# x
'c
1Î3
Ê Aœ œ Š8 †
1Î3
È3 #
1Î$
a8 cos x sec# xb dx œ [8 sin x tan x]1Î$
È3‹ Š8 †
È3 #
È3‹ œ 6È3
27. f(y) œ Èy, g(y) œ 2 y, c œ 1, d œ 2
'
Ê Aœ œ
'
1
2
d
c
[f(y) g(y)] dy œ
'
2
1
Èy (2 y)‘ dy
ˆÈy 2 y‰ dy œ ’ 23 y$Î# 2y
œ Š 43 È2 4 2‹ ˆ 23 2 "# ‰ œ
4 3
# y# # “"
È2
7 6
œ
8 È 2 7 6
28. f(y) œ 6 y, g(y) œ y# , c œ 1, d œ 2 Ê Aœ
'
œ ’6y
y# #
œ4
c
7 3
d
[f(y) g(y)] dy œ
" #
# y$ 3 “"
œ
'
2
1
a6 y y# b dy
œ ˆ12 2 83 ‰ ˆ6
24143 6
œ
" #
3" ‰
13 6
29. f(x) œ x$ 3x# œ x# (x 3) Ê f w (x) œ 3x# 6x œ 3x(x 2) Ê f w œ ± ± ! # Ê f(0) œ 0 is a maximum and f(2) œ 4 is a minimum. A œ ‰ œ ˆ 81 4 27 œ 30. A œ
'
a
0
4 3
a# 6
"# ‰ œ
'
a
&Î$
A# œ
" y# # “!
œ
0
(6 8 3) œ
" 10
'
0
1
a
x# # “0
$ !
œ a# 34 Èa † aÈa
a# 6
ˆy#Î$ y‰ dy
; the area below the x-axis is
'c ˆy#Î$ y‰ dy œ ’ 3y5 0
%
ax$ 3x# b dx œ ’ x4 x$ “
ˆa 2Èa x"Î# x‰ dx œ ’ax 43 Èa x$Î#
31. The area above the x-axis is A" œ œ ’ 3y5
0
3
27 4
ˆa"Î# x"Î# ‰# dx œ
œ a# ˆ1
'
&Î$
1
Ê the total area is A" A# œ
! y# # “ "
œ
11 10
6 5
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
a# #
Chapter 5 Practice Exercises
'
32. A œ
1Î4
0
'
31Î2
51Î4
(cos x sin x) dx
'
51Î4
1Î4
(sin x cos x) dx 1Î%
(cos x sin x) dx œ [sin x cos x]! &1Î%
$1Î#
[ cos x sin x]1Î% [sin x cos x]&1Î% œ ’Š
È2 #
È2 # ‹
(0 1)“ ’Š
’(1 0) Š
33. y œ x# 34. y œ
'
x
0
'
x
" 1 t
È2 #
dt Ê
dy dx
È2 # ‹“
xœ0 Ê yœ
'
0
0
È2 # ‹
È2 #
Š
œ
8È 2 #
2 œ 4È2 2
" x
Ê
d# y dx#
œ 2x
ˆ1 2Èsec t‰ dt Ê
È2 #
œ2
" x#
; y(1) œ 1
œ 1 2Èsec x Ê
dy dx
d# y dx#
ˆ1 2Èsec t‰ dt œ 0 and x œ 0 Ê
dy dx
36. y œ
'c È2 sin# t dt 2 so that dydx œ È2 sin# x; x œ 1
sin t t
dt 3 Ê
dy dx
œ
;xœ5 Ê yœ
sin x x
5
5
sin t t
1
1
" t
dt œ 1 and yw (1) œ 2 1 œ 3
œ 1 2Èsec 0 œ 3
'
x
'
œ 2 ˆ "# ‰ (sec x)"Î# (sec x tan x) œ Èsec x (tan x);
35. y œ
5
'
È2 # ‹“
dt 3 œ 3
x
Ê yœ
1
'cc È2 sin# t dt 2 œ 2 1
1
37. Let u œ cos x Ê du œ sin x dx Ê du œ sin x dx
' 2(cos x)"Î# sin x dx œ ' 2u"Î# ( du) œ 2 ' u"Î# du œ 2 Š u"Î#" ‹ C œ 4u"Î# C #
œ 4(cos x)"Î# C 38. Let u œ tan x Ê du œ sec# x dx
' (tan x)$Î# sec# x dx œ ' u$Î# du œ ˆu"Î#"‰ C œ 2u"Î# C œ (tanx)2 "Î# C #
39. Let u œ 2) 1 Ê du œ 2 d) Ê
" #
du œ d)
' [2) 1 2 cos (2) 1)] d) œ ' (u 2 cos u) ˆ "# du‰ œ u4
#
œ )# ) sin (2) 1) C, where C œ C" 40. Let u œ 2) 1 Ê du œ 2 d) Ê
'Š œ
41.
42.
" #
" È 2 ) 1
" #
2 sec# (#) 1)‹ d) œ
" 4
sin u C" œ
(2)1)# 4
sin (2) 1) C"
is still an arbitrary constant
du œ d)
' Š È"u 2 sec# u‹ ˆ #" du‰ œ #" ' ˆu"Î# 2 sec# u‰ du
"Î#
Š u " ‹ "# (2 tan u) C œ u"Î# tan u C œ (2) 1)"Î# tan (2) 1) C #
' ˆt 2t ‰ ˆt 2t ‰ dt œ ' ˆt# t4 ‰ dt œ ' at# 4t# b dt œ t3$ 4 Š t"1 ‹ C œ t3$ 4t C #
t# ' (t1)t%#1 dt œ ' t#t%2t dt œ ' ˆ t"# t2$ ‰ dt œ ' at# 2t$ b dt œ (t"1) 2 Š # ‹ C œ "t t"# C
43. Let u œ #t$Î# Ê du œ $Èt dt Ê "$ du œ Èt dt
' Èt sin ˆ#t$Î# ‰dt œ "$ ' sin u du œ "$ cos u C œ "$ cosˆ#t$Î# ‰ C Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
349
350
Chapter 5 Integration
44. Let u œ " sec ) Ê du œ sec ) tan) d) Ê
' sec ) tan) È" sec ) d) œ ' u"Î# du œ #$ u$Î# C
œ #$ a" sec )b$Î# C
'c a3x# 4x 7b dx œ cx$ 2x# 7xd "" œ c1$ 2(1)# 7(1)d c(1)$ 2(1)# 7(1)d œ 6 (10) œ 16 1
45.
1
46.
'
47.
'
48.
'
49.
'
1
0
"
a8s$ 12s# 5b ds œ c2s% 4s$ 5sd ! œ c2(1)% 4(1)$ 5(1)d 0 œ 3
2
4 # 1 v 27
1
'
dv œ
2
#
4v# dv œ c4v" d " œ ˆ #4 ‰ ˆ 14 ‰ œ 2
1
#(
x%Î$ dx œ 3x"Î$ ‘ " œ 3(27)"Î$ ˆ3(1)"Î$ ‰ œ 3 ˆ "3 ‰ 3(1) œ 2
4
dt 1 tÈt
œ
'
4
œ
dt
t$Î#
1
'
4
1
%
50. Let x œ 1 Èu Ê dx œ
'
4
ˆ1 Èu‰"Î# Èu
1
du œ
'
3
2
2 È4
t$Î# dt œ 2t"Î# ‘ " œ " #
u"Î# du Ê 2 dx œ
du Èu
(2) È1
œ1
; u œ 1 Ê x œ 2, u œ 4 Ê x œ 3
$
x"Î# (2 dx) œ 2 ˆ 23 ‰ x$Î# ‘ # œ
4 3
ˆ3$Î# ‰ 43 ˆ2$Î# ‰ œ 4È3 83 È2 œ
4 3
Š3È3 2È2‹
51. Let u œ 2x 1 Ê du œ 2 dx Ê 18 du œ 36 dx; x œ 0 Ê u œ 1, x œ 1 Ê u œ 3
'
1
36 dx $ 0 (2x1)
œ
'
3
# $
$
9 ‘ ˆ 9 ‰ ˆ 9 ‰ 18u$ du œ ’ "8u 2 “ œ u # " œ 3 # 1 # œ 8 "
1
52. Let u œ 7 5r Ê du œ 5 dr Ê "5 du œ dr; r œ 0 Ê u œ 7, r œ 1 Ê u œ 2
'
1
dr $ È (7 5r)#
0
œ
'
'
1
(7 5r)#Î$ dr œ
0
2
7
#
u#Î$ ˆ 5" du‰ œ 5" 3u"Î$ ‘ ( œ
53. Let u œ 1 x#Î$ Ê du œ 23 x"Î$ dx Ê 3# du œ x"Î$ dx; x œ x œ 1 Ê u œ 1 1#Î$ œ 0
'
1
1Î8
œ
x"Î$ ˆ1 x#Î$ ‰
$Î#
'
dx œ
0
1Î2
0
x$ a1 9x% b
" ˆ 25 ‰ œ 18 16
"Î#
$Î#
dx œ
'
0
1
sin# 5r dr œ
56. Let u œ 4t
'
1Î%
0
œ
1 8
1 4
$Î%
" 36
#Î$
œ
3 4
,
! &Î# œ 35 u&Î# ‘ $Î% œ 35 (0)&Î# ˆ 35 ‰ ˆ 34 ‰
'
51
0
" 16
" 16
" 5
"Î# #
1 8
' Î acos 1 4
"
%
Ê u œ 1 9 ˆ #" ‰ œ
25 16
#&Î"'
" "Î# ‘ œ 18 u "
du œ dr; r œ 0 Ê u œ 0, r œ 1 Ê u œ 51
Ê du œ 4 dt Ê
œ
#&Î"'
" #
" 90
asin# ub ˆ "5 du‰ œ
31Î4
du œ x$ dx; x œ 0 Ê u œ 1, x œ
" " u$Î# ˆ 36 du‰ œ ’ 36 Š u " ‹“
" ˆ 18 (1)"Î# ‰ œ
cos# ˆ4t 14 ‰ dt œ
25Î16
1
55. Let u œ 5r Ê du œ 5 dr Ê
'
!
#
Ê u œ 1 ˆ 8" ‰
27È3 160
54. Let u œ 1 9x% Ê du œ 36x$ dx Ê
'
&Î#
u$Î# ˆ #3 du‰ œ ’ˆ 32 ‰ Š u 5 ‹“
3Î4
" 8
Š $È7 $È2‹
3 5
" 4
#
" 5
u2
sin 2u ‘ &1 4 !
œ ˆ 1#
sin 101 ‰ #0
du œ dt; t œ 0 Ê u œ 14 , t œ ub ˆ "4 du‰ œ
" 4
u2
sin 2u ‘ $1Î% 4 1Î%
œ
ˆ0 1 4 " 4
sin 0 ‰ 20
Ê uœ Š 381
œ
1 #
31 4
sin ˆ 3#1 ‰ ‹ 4
4" Š 18
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
sin ˆ 1# ‰ ‹ 4
Chapter 5 Practice Exercises
'
1Î$
57.
'
31Î4
58.
0
1Î$
sec# ) d) œ [tan )]!
1Î4
31
1
1 3
tan 0 œ È3
$1Î%
csc# x dx œ [cot x]1Î% œ ˆ cot
59. Let u œ
'
œ tan
cot#
" 6
Ê du œ
x 6
dx œ
x 6
'
1Î6
31 ‰ 4
ˆ cot 14 ‰ œ 2
dx Ê 6 du œ dx; x œ 1 Ê u œ 16 , x œ 31 Ê u œ
1Î2
351
'
6 cot# u du œ 6
1Î2
1 # 1Î#
acsc# u 1b du œ [6(cot u u)]1Î' œ 6 ˆ cot
1Î6
1 #
1# ‰ 6 ˆcot
1 6
16 ‰
œ 6È3 21 60. Let u œ
'
1
0
tan#
) 3 ) 3
œ 3 tan
" 3
Ê du œ d) œ 1 3
'
0
1
d) Ê 3 du œ d); ) œ 0 Ê u œ 0, ) œ 1 Ê u œ ) 3
ˆsec#
1‰ d) œ
'
1Î3
0
1 3 1Î$
3 asec# u 1b du œ [3 tan u 3u]!
3 ˆ 13 ‰‘ (3 tan 0 0) œ 3È3 1
'c
sec x tan x dx œ [sec x]!1Î$ œ sec 0 sec ˆ 13 ‰ œ 1 2 œ 1
'
csc z cot z dz œ [csc z]1Î% œ ˆ csc
0
61.
62.
1Î3
31Î4
1Î4
$1Î%
31 ‰ 4
ˆ csc 14 ‰ œ È2 È2 œ 0
63. Let u œ sin x Ê du œ cos x dx; x œ 0 Ê u œ 0, x œ
'
1Î2
0
'
5(sin x)$Î# cos x dx œ
1
0
1 #
Ê uœ1
"
"
5u$Î# du œ 5 ˆ 25 ‰ u&Î# ‘ ! œ 2u&Î# ‘ ! œ 2(1)&Î# 2(0)&Î# œ 2
64. Let u œ 1 x# Ê du œ 2x dx Ê du œ 2x dx; x œ 1 Ê u œ 0, x œ 1 Ê u œ 0
'c
1 1
'
2x sin a1 x# b dx œ
0
sin u du œ 0
0
" 3
65. Let u œ sin 3x Ê du œ 3 cos 3x dx Ê œ 1
'c ÎÎ
1 2
15 sin% 3x cos 3x dx œ
1 2
du œ cos 3x dx; x œ 1# Ê u œ sin ˆ 3#1 ‰ œ 1, x œ
1 #
& & ' 15u% ˆ "3 du‰ œ ' 5u% du œ cu& d " " œ (1) (1) œ 2 1
1
1
1
66. Let u œ cos ˆ x# ‰ Ê du œ "# sin ˆ x# ‰ dx Ê 2 du œ sin ˆ x# ‰ dx; x œ 0 Ê u œ cos ˆ 0# ‰ œ 1, x œ œ
'
" #
21Î3
cos% ˆ x# ‰ sin ˆ x# ‰ dx œ
0
'
1
1Î2
$
u% (2 du) œ ’2 Š u3 ‹“
67. Let u œ 1 3 sin# x Ê du œ 6 sin x cos x dx Ê Ê u œ 1 3 sin#
'
1Î2
0
3 sin x cos x È1 3 sin# x
1 #
œ4
dx œ
'
4
" Èu
1
ˆ #" du‰ œ
'
4
1
" #
'
0
1Î4
1 4
Ê u œ 1 7 tan
sec# x (1 7 tan x)#Î$
dx œ
'
1
1 4 8
" #
"Î# "
œ
2 3
ˆ "# ‰$ 32 (1)$ œ
2 3
(8 1) œ
du œ 3 sin x cos x dx; x œ 0 Ê u œ 1, x œ "Î#
%
#
"
21 3
21
Ê u œ cos Š #3 ‹
14 3
1 #
%
u"Î# du œ ’ 2" Š u " ‹“ œ u"Î# ‘ " œ 4"Î# 1"Î# œ 1 " 7
68. Let u œ 1 7 tan x Ê du œ 7 sec# x dx Ê xœ
Ê u œ sin ˆ 3#1 ‰
du œ sec# x dx; x œ 0 Ê u œ 1 7 tan 0 œ 1,
œ8 " u#Î$
ˆ 7" du‰ œ
'
1
8
" 7
"Î$
)
3
"
)
u#Î$ du œ ’ 7" Š u " ‹“ œ 37 u"Î$ ‘ " œ
3 7
(8)"Î$ 37 (1)"Î$ œ
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
3 7
352
Chapter 5 Integration
69. Let u œ sec ) Ê du œ sec ) tan ) d); ) œ 0 Ê u œ sec 0 œ 1, ) œ
'
1Î3
0
tan ) È2 sec )
" È2
œ
'
d) œ
"Î#
#
#
"
'
1Î3
0
’ ˆu " ‰ “ œ ’
1Î3
sec ) tan ) sec ) tan ) d) œ È2 (sec ))$Î# sec ) È2 sec ) 0 # 2 2 2 È2u “ œ È2(2) Š È2(1) ‹ œ " cos Èt 2È t
70. Let u œ sin Èt Ê du œ ˆcos Èt‰ ˆ "# t"Î# ‰ dt œ #
1 4
tœ
'
1 Î4 #
Ê u œ sin
71. (a) av(f) œ
'
" 1 (1)
'c
" k (k)
(b) av(f) œ " #k
œ
1
" 1Î2 Èu
1 1
'c
k k
(2 du) œ 2
73. favw œ
'
a
b
" È2 u$Î#
du œ
" È2
'
1
2
œ2 u$Î# du
È2 1 dt Ê 2 du œ
cos Èt Èt
dt; t œ
1# 36
Ê u œ sin
1 6
œ
" #
,
"
"
" #k
’ mx2 bx“
"
#
’ mx2 bx“
(2bk) œ b " 30
1
1 3
u"Î# du œ 4Èu‘ "Î# œ 4È1 4É #" œ 2 Š2 È2‹ #
(mx b) dx œ
3
0
3
0
0
" ba
1
1Î2
" #
a
(b) yav
'
(mx b) dx œ
' È3x dx œ "3 ' œ a " 0 ' Èax dx œ "a '
72. (a) yav œ
'
Ê u œ sec
œ1
dt œ
cos Èt Ét sin Èt
1# Î36
1 #
d) œ
2
1 3
È3 x"Î# dx œ
a
Èa x"Î# dx œ
0
" b a
Èaxf w (x) dx œ
[f(x)]ab œ
" b a
k
ck
#
#
m(1) b(1)‹“ œ ’Š m(1) 2 b(1)‹ Š #
œ
" #
œ
" #k
#
" #
(2b) œ b
#
m(k) b(k)‹“ ’Š m(k) 2 b(k)‹ Š #
È3 3
23 x$Î# ‘ $ œ !
È3 3
23 (3)$Î# 23 (0)$Î# ‘ œ
È3 3
Š2È3‹ œ 2
Èa a
23 x$Î# ‘ a œ !
Èa a
ˆ 23 (a)$Î# 23 (0)$Î# ‰ œ
Èa a
ˆ 32 aÈa‰ œ
[f(b) f(a)] œ
f(b) f(a) ba
2 3
a
so the average value of f w over [aß b] is the
slope of the secant line joining the points (aß f(a)) and (bß f(b)), which is the average rate of change of f over [aß b]. 74. Yes, because the average value of f on [aß b] is and the average value of the function is
" #
'
a
" ba
'
a
b
f(x) dx. If the length of the interval is 2, then b a œ 2
b
f(x) dx.
75. We want to evaluate " $'& !
'
$'&
!
f(x) dx œ
" $'&
'
$'&
!
#1 Œ$(sin” $'& ax "!"b• #&dx œ
#1 Notice that the period of y œ sin” $'& ax "!"b• is
length 365. Thus the value of
76.
" '(&#!
œ
'#
'(& !
$( $'&
'
$'&
!
" '&& Œ”)Þ#(a'(&b
#'a'(&b #†"!&
#1 $'&
"Þ)(a'(&b $†"!&
$
'
!
$'&
#1 sin” $'& ax "!"b•dx
#& $'&
'
$'&
!
dx
œ $'& and that we are integrating this function over an iterval of
#1 ax "!"b•dx sin” $'&
a)Þ#( "!& a#'T "Þ)(T# bbdT œ #
#1
$( $'&
" '&& ”)Þ#(T
• ”)Þ#(a#!b
#& $'&
#'T# #†"!&
#'a#!b #†"!&
#
'
!
$'&
dx is
"Þ)(T$ $†"!& • "Þ)(a#!b $†"!&
$
$( $'&
†!
#& $'&
† $'& œ #&.
'(& #!
• ¸
" '&& a$(#%Þ%%
"'&Þ%!b
œ &Þ%$ œ the average value of Cv on [20, 675]. To find the temperature T at which Cv œ &Þ%$, solve &Þ%$ œ )Þ#( "!& a#'T "Þ)(T# b for T. We obtain "Þ)(T# #'T #)%!!! œ ! ÊTœ
#' „ Éa#'b# %a"Þ)(ba#)%!!!b È#"#%**' œ #' „ $Þ(% . #a"Þ)(b ‰
So T œ $)#Þ)# or T œ $*'Þ(#. Only T œ $*'Þ(# lies in the
interval [20, 675], so T œ $*'Þ(# C. 77.
dy dx
œ È# cos$ x
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 5 Practice Exercises 78.
dy dx
œ È# cos$ a(x# b †
79.
dy dx
œ
d dx Œ
80.
dy dx
œ
d dx Œ
d # dx a(x b
œ "%xÈ# cos$ a(x# b
' x $ ' t dt œ $'x %
1
353
%
'sec# x t " " dt œ dxd Œ'#sec x t " " dt œ sec "x " dxd asec xb œ sec" xsectan xx #
#
#
#
81. Yes. The function f, being differentiable on [aß b], is then continuous on [aß b]. The Fundamental Theorem of Calculus says that every continuous function on [aß b] is the derivative of a function on [aß b]. 82. The second part of the Fundamental Theorem of Calculus states that if F(x) is an antiderivative of f(x) on [aß b], then
'
0
1
'
b
a
f(x) dx œ F(b) F(a). In particular, if F(x) is an antiderivaitve of È1 x% on [0ß 1], then
È1 x% dx œ F(1) F(0).
83. y œ
' x È1 t# dt œ ' x È1 t# dt
84. y œ
'
1
1
0
" # cos x 1 t
dt œ
'
0
cos x
" 1 t#
dt Ê
Ê
dy dx
œ
d dx
dy dx
œ
d dx
”
' x È1 t# dt• œ dxd ”' x È1 t# dt• œ È1 x#
”
'
1
cos x
0
" d ‰ ˆ dx œ ˆ 1 cos (cos x)‰ œ ˆ sin"# x ‰ ( sin x) œ #x
1
" 1 t#
" sin x
d dt• œ dx ”
'
0
cos x
" 1 t#
dt•
œ csc x
85. We estimate the area A using midpoints of the vertical intervals, and we will estimate the width of the parking lot on each interval by averaging the widths at top and bottom. This gives the estimate A ¸ "& † ˆ ! # $' $' # &% &% # &" &" #%*Þ& %*Þ&# &% &% #'%Þ% '%Þ% # '(Þ& '(Þ&# %# ‰
A ¸ &*'" ft# . The cost is Area † ($2.10/ft# ) ¸ a5961 ft# b a$2.10/ft# b œ $12,518.10 Ê the job cannot be done for $11,000.
86. (a) Before the chute opens for A, a œ 32 ft/sec# . Since the helicopter is hovering, v! œ 0 ft/sec
Ê v œ ' 32 dt œ 32t v! œ 32t. Then s! œ 6400 ft Ê s œ ' 32t dt œ 16t# s! œ 16t# 6400.
At t œ 4 sec, s œ 16(4)# 6400 œ 6144 ft when A's chute opens;
(b) For B, s! œ 7000 ft, v! œ 0, a œ 32 ft/sec# Ê v œ ' 32 dt œ 32t v! œ 32t Ê s œ ' 32t dt œ 16t# s! œ 16t# 7000. At t œ 13 sec, s œ 16(13)# 7000 œ 4296 ft when B's chute opens;
(c) After the chutes open, v œ 16 ft/sec Ê s œ ' 16 dt œ 16t s! . For A, s! œ 6144 ft and for B,
s! œ 4296 ft. Therefore, for A, s œ 16t 6144 and for B, s œ 16t 4296. When they hit the ground, 4296 s œ 0 Ê for A, 0 œ 16t 6144 Ê t œ 6144 16 œ 384 seconds, and for B, 0 œ 16t 4296 Ê t œ 16 œ 268.5 seconds to hit the ground after the chutes open. Since B's chute opens 58 seconds after A's opens Ê B hits the ground first. 87. av(I) œ œ
" 30
" 30
'
30
0
(1200 40t) dt œ
" 30
$!
c1200t 20t# d! œ
" 30
ca(1200(30) 20(30)# b a1200(0) 20(0)# bd
(18,000) œ 600; Average Daily Holding Cost œ (600)($0.03) œ $18
88. av(I) œ
" 14
'
0
14
(600 600t) dt œ
" 14
"%
c600t 300t# d! œ
" 14
c600(14) 300(14)# 0d œ 4800; Average Daily
Holding Cost œ (4800)($0.04) œ $192
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
354
Chapter 5 Integration
'
" 30
89. av(I) œ
30
#
0
$
" 30
Š450 t# ‹ dt œ
’450t t6 “
œ (300)($0.02) œ $6
œ
" 60
'
" 60
90. av(I) œ
60
0
40È15 3
’600(60)
(60)
$Î#
'
" 60
Š600 20È15t‹ dt œ
0
" 60
0“ œ
60
$! !
" 30
œ
30$ 6
’450(30)
0“ œ 300; Average Daily Holding Cost
Š600 20È15 t"Î# ‹ dt œ
" 60
’600t 20È15 ˆ 23 ‰ t$Î# “
'! !
ˆ36,000 ˆ 320 ‰ 15# ‰ œ 200; Average Daily Holding Cost 3
œ (200)($0.005) œ $1.00 CHAPTER 5 ADDITIONAL AND ADVANCED EXERCISES
'
1. (a) Yes, because
1
0
(b) No. For example, 4È 2 3
œ
'
1
0
'
" 7
f(x) dx œ
1
0
" 7
7f(x) dx œ
(7) œ 1
'
"
8x dx œ c4x# d ! œ 4, but
1
0
"
È8x dx œ ’2È2 Š x$Î# œ 3 ‹“
2
2
5
5
5
2
2
2
2
2
œ432œ9
'c f(x) dx œ 4 3 œ 7 2 œ 'c g(x) dx
(c) False:
5
5
2
2
œ
'
x
0
sin ax a sin ax a
'
0
f(t) cos at dt
'
d
Πdx
œ cos ax
x
0
'
x
0
f(x) dx
5
2
5
2
x
0
0
x
x
0
0
'
f(t) cos at dt
sin ax a
x
0
f(t) sin at dt
(f(x) cos ax) sin ax
x
cos ax a
'
x
0
d
Πdx
'
0
x
f(t) sin at dt
f(t) sin at dt
cos ax a
(f(x) sin ax)
x
0 x
#
#
0
x
0
'
d (sin ax) Πdx
a cos ax
'
0
x
0
x
0
f(t) sin at dt œ a sin ax
4. x œ
'
" #
x
0
y
" 0 È1 4t#
Ê 1œ œ
'
'
f(t) cos at dt
dt Ê
" È14y#
a1 4y# b
'
0
x
d dx
Š dy dx ‹ Ê
"Î#
(x) œ dy dx
0
x
'
d dx
0
'
0
y
x
'
'
" È1 4t#
dy 4y Š dx ‹
È1 4y#
0
x
0
x
f(t) cos at dt a cos ax
'
0
x
f(t) sin at dt f(x).
f(t) cos at dt f(x)
f(t) sin at dt œ f(x). Note also that yw (0) œ y(0) œ 0.
dt œ
œ È1 4y# . Then
(8y) Š dy dx ‹ œ
f(t) sin at dt
f(t) cos at dt (cos ax)f(x) cos ax
f(t) sin at dt a sin ax
cos ax a
x
0
f(t) sin at dt (sin ax)f(x) sin ax œ a sin ax
Therefore, yww a# y œ a cos ax a# Œ sinaax
2
2
x
" a
f(t) cos at dt sin ax
œ cos ax
dy dx
5
5
' f(t) cos at dt sin ax ' f(t) sin at dt. Next, d y ' f(t) cos at dt (cos ax) Œ dxd ' f(t) cos at dt a cos ax ' dx œ a sin ax Ê
'c g(x) dx
' f(t) sin ax cos at dt "a ' f(t) cos ax sin at dt cos ax ' ' f(t) cos at dt f(t) sin at dt Ê dy a dx œ cos ax Œ
f(t) sin a(x t) dt œ x
2
5
'c [f(x) g(x)] dx 0 Ê 'c [g(x) f(x)] dx 0. Ê ' [g(x) f(x)] dx 0 which is a contradiction. c
Ê
On the other hand, f(x) Ÿ g(x) Ê [g(x) f(x)] 0 " a
ˆ1$Î# 0$Î# ‰
5
5
(b) True:
4È 2 3
Á È4
' f(x) dx œ ' f(x) dx œ 3 'c [f(x) g(x)] dx œ 'c f(x) dx 'c g(x) dx œ 'c f(x) dx '
2. (a) True:
3. y œ
!
#
œ
'
d dy
”
d# y dx#
œ
4y ˆÈ1 4y# ‰ È1 4y#
y
0
" È1 4t#
d dx
ˆÈ1 4y# ‰ œ
œ 4y. Thus
dt• Š dy dx ‹ from the chain rule
d# y dx#
d dy
ˆÈ1 4y# ‰ Š dy dx ‹
œ 4y, and the constant of
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 5 Additional and Advanced Exercises
355
proportionality is 4.
'
5. (a)
x#
f(t) dt œ x cos 1x Ê
0
cos 1x 1x sin 1x . 2x
Ê f ax# b œ
'
(b)
f(x)
0
d dx
$
t# dt œ ’ t3 “
f(x)
!
œ
" 3
'
x#
f(t) dt œ cos 1x 1x sin 1x Ê f ax# b (2x) œ cos 1x 1x sin 1x
0
cos 21 21 sin 21 4
Thus, x œ 2 Ê f(4) œ " 3
(f(x))$ Ê
" 4
œ
(f(x))$ œ x cos 1x Ê (f(x))$ œ 3x cos 1x Ê f(x) œ $È3x cos 1x
Ê f(4) œ $È3(4) cos 41 œ $È12 6.
'
a
f(x) dx œ
0
a# #
a #
Ê f(a) œ Fw (a) œ a 7.
'
b
1
1 #
sin a " #
cos a. Let F(a) œ
sin a
a #
1 #
cos a
f(x) dx œ Èb# 1 È2 Ê f(b) œ
d db
'
a
0
f(t) dt Ê f(a) œ Fw (a). Now F(a) œ
sin a Ê f ˆ 1# ‰ œ
'
b
1
f(x) dx œ
" #
side of the equation is: œ
d dx
œ
'
0
'
”x
0
x
f(u) du•
'
d dx
”
d dx
'
x
0
x
0
f(u)(x u) du• œ
u f(u) du œ
'
x
0
' ”'
dy dx
d dx
0
'
x
0
0
u
" #
1 #
sin
"Î#
'
0
x
d dx
(2b) œ
f(t) dt• du• œ
f(u) x du
d f(u) du x ” dx
'
'
0
cos
1 #
1 #
a #
sin a
sin
1 #
Ê f(x) œ
b È b# 1
œ
1 #
1 #
cos a
" #
1 #
œ
" #
x È x# 1
x
f(t) dt; the derivative of the right
x
0
u f(u) du
f(u) du• xf(x) œ
'
0
x
f(u) du xf(x) xf(x)
x
f(u) du. Since each side has the same derivative, they differ by a constant, and since both sides equal 0
when x œ 0, the constant must be 0. Therefore,
9.
”
ab# 1b
x
d dx
8. The derivative of the left side of the equation is:
1 #
ˆ 1# ‰ #
a# #
' ”' x
0
0
u
f(t) dt• du œ
'
0
x
f(u)(x u) du.
œ 3x# 2 Ê y œ ' a3x# 2b dx œ x$ 2x C. Then (1ß 1) on the curve Ê 1$ 2(1) C œ 1 Ê C œ 4
Ê y œ x$ 2x 4 10. The acceleration due to gravity downward is 32 ft/sec# Ê v œ ' 32 dt œ 32t v! , where v! is the initial
velocity Ê v œ 32t 32 Ê s œ ' (32t 32) dt œ 16t# 32t C. If the release point, at t œ !, is s œ 0, then C œ 0 Ê s œ 16t# 32t. Then s œ 17 Ê 17 œ 16t# 32t Ê 16t# 32t 17 œ 0. The discriminant of this quadratic equation is 64 which says there is no real time when s œ 17 ft. You had better duck.
11.
'c f(x) dx œ 'c x#Î$ dx ' œ œ œ
12.
3
0
8
8
3
4 dx
0
35 x&Î$ ‘ ! [4x]!$ ) ˆ0 35 (8)&Î$ ‰ (4(3) 36 5
0) œ
'c f(x) dx œ 'c Èx dx ' 3
0
4
4
!
3
0
$
œ 23 (x)$Î# ‘ % ’ x3 4x“
96 5
12
ax# 4b dx $ !
$
œ 0 ˆ 23 (4)$Î# ‰‘ ’ Š 33 4(3)‹ 0 “ œ
16 3
3œ
7 3
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
356 13.
Chapter 5 Integration
'
2
0
'
g(t) dt œ
1
t dt
0
"
#
'
2
1
sin 1t dt
#
œ ’ t2 “ 1" cos 1t‘ " !
œ ˆ "# 0‰ 1" cos 21 ˆ 1" cos 1‰‘ " #
œ
14.
'
2
0
2 1
h(z) dz œ
'
1
0
'
È1 z dz
1
2
(7z 6)"Î$ dz
" # 3 œ 23 (1 z)$Î# ‘ ! 14 (7z 6)#Î$ ‘ " œ 23 (1 1)$Î# ˆ 23 (1 0)$Î# ‰‘ 3 14 (7(2) 6)#Î$ 6 3 ‰ 55 œ ˆ 7 14 œ 42
3 14
(7(1) 6)#Î$ ‘
2 3
'c f(x) dx œ 'cc dx 'c a1 x# b dx ' 2
15.
1
2
1
2
1
"
x$ 3 “ "
œ [x]" # ’x
1$ 3‹
16.
ˆ 23 ‰ 4 2 œ
2 3
'c h(r) dr œ 'c r dr ' 2
0
1
1
#
œ ’ r2 “
!
"
2 3
Š Š1
1œ
a1 r# b dr
" ba
1$ 3‹
'
2
1
7 6
'
b
a
f(x) dx œ
" #0
'
2
0
f(x) dx œ
#
’Š 1# 0‹ Š 2# 2‹ Š 1# 1‹“ œ
'
20. f(x) œ
'
x
" t
1/x
" ba
sin x
" " t 1 t# " sin x
'
a
'ÈÈ sin t# dt
22. f(x) œ
'
y
x
xb3
f(x) dx œ
y
" x
" 30
'
3
0
" #
”
'
1
0
x dx
'
2
1
(x 1) dx• œ
" #
#
"
#
’ x2 “ #" ’ x2 x“ !
# "
" #
f(x) dx œ
" 3
”
'
dx ‰ d ˆ " ‰‰ ˆ dx Š "" ‹ ˆ dx œ x x
0
" x
1
dx
'
1
2
0 dx
x ˆ x"# ‰ œ
" x
'
3
2
dx• œ
" x
œ
#
#
" 3
[1 0 0 3 2] œ
2 3
2 x
" d " d ‰ ˆ dx ‰ ˆ dx dt Ê f w (x) œ ˆ 1 sin (sin x)‰ ˆ 1 cos (cos x)‰ œ #x #x
21. g(y) œ
2
b
dt Ê f w (x) œ
cos x
" cos x
dr
0‹ a2 1b
#
18. Ave. value œ
œ
13 3
"
#
19. f(x) œ
’2(2) 2(1)“
!
(1)# # ‹
17. Ave. value œ " #
(1)$ 3 ‹•
Š1
’r r3 “ [r]#"
œ "#
œ
1
0
$
œ Š0
2 dx
[2x]#"
œ a1 (2)b ”Š1 œ1
2
1
cos x cos# x
sin x sin# x
d ˆ d ˆ Èy‰‹ œ Ê gw (y) œ Šsin ˆ2Èy‰ ‹ Š dy 2Èy‰‹ Šsin ˆÈy‰ ‹ Š dy
sin 4y Èy
sin y 2È y
d ‰ t(5 t) dt Ê f w (x) œ (x 3)(& (x 3)) ˆ dx (x 3)‰ x(5 x) ˆ dx dx œ (x 3)(2 x) x(5 x)
œ 6 x x# 5x x# œ 6 6x. Thus f w (x) œ 0 Ê 6 6x œ 0 Ê x œ 1. Also, f ww (x) œ 6 ! Ê x œ 1 gives a
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 5 Additional and Advanced Exercises maximum. 23. Let f(x) œ x& on [0ß 1]. Partition [0ß 1] into n subintervals with ?x œ
10 n
œ "n . Then "n , n2 , á ,
_
n n
&
are the
right-hand endpoints of the subintervals. Since f is increasing on [0ß 1], U œ ! Š nj ‹ ˆ "n ‰ is the upper sum for
_
œ
j 1
&
! Š j ‹ ˆ"‰ œ f(x) œ x& on [0ß 1] Ê n lim n n Ä_ jœ1 œ
'
1
'
"
x& dx œ ’ x6 “ œ !
0
& lim " ’ˆ n" ‰ nÄ_ n
ˆ n2 ‰&
&
á ˆ nn ‰ “ œ n lim ’1 Ä_
&
2& á n& “ n'
" 6
24. Let f(x) œ x$ on [0ß 1]. Partition [0ß 1] into n subintervals with ?x œ
10 n
œ "n . Then "n , n2 , á ,
_
n n
$
are the
right-hand endpoints of the subintervals. Since f is increasing on [0ß 1], U œ ! Š nj ‹ ˆ "n ‰ is the upper sum for
_
œ
j 1
$
! Š j ‹ ˆ " ‰ œ lim f(x) œ x$ on [0ß 1] Ê n lim n n Ä_ nÄ_ jœ1 œ
'
0
1
%
"
x$ dx œ ’ x4 “ œ !
" n
$ $ $ ’ˆ n" ‰ ˆ n2 ‰ á ˆ nn ‰ “ œ n lim ’1 Ä_
$
2$ á n$ “ n%
" 4
25. Let y œ f(x) on [0ß 1]. Partition [0ß 1] into n subintervals with ?x œ
10 n
œ "n . Then "n , 2n , á ,
_
n n
are the
right-hand endpoints of the subintervals. Since f is continuous on [!ß 1], ! f Š nj ‹ ˆ "n ‰ is a Riemann sum of
œ
j 1
_
! f Š j ‹ ˆ " ‰ œ lim y œ f(x) on [0ß 1] Ê n lim n n Ä_ nÄ_ jœ1
" n
'
1
" 26. (a) n lim [2 4 6 á 2n] œ n lim Ä _ n# Ä_ on [0ß 1] (see Exercise 25)
" n
n2
" (b) n lim c1"& 2"& á n"& d œ n lim Ä _ n"' Ä_ "& f(x) œ x on [0ß 1] (see Exercise 25)
" n
"& "& "& ’ˆ 1n ‰ ˆ 2n ‰ á ˆ nn ‰ “ œ
'
'
f ˆ n" ‰ f ˆ n2 ‰ á f ˆ nn ‰‘ œ
4 n
6 n
á
œ
2n ‘ n
0
1
0
f(x) dx
"
2x dx œ cx# d! œ 1, where f(x) œ 2x
'
0
1
"'
"
" 16 ,
x"& dx œ ’ x16 “ œ !
where
1
" " (c) n lim sin 1n sin 2n1 á sin nn1 ‘ œ sin n1 dx œ 1" cos 1x‘ ! œ 1" cos 1 ˆ 1" cos 0‰ Ä_ n 0 œ 12 , where f(x) œ sin 1x on [0ß 1] (see Exercise 25) " (d) n lim c1"& 2"& á n"& d œ Šn lim Ä _ n"( Ä_ " ‰ ˆ œ 0 16 œ 0 (see part (b) above) " n"&
(e) n lim Ä_
c1"& 2"& á n"& d œ n lim Ä_
œ Šn lim n‹ Šn lim Ä_ Ä_
" n"'
" " n ‹ Šn lim Ä _ n"'
n n"'
c1"& 2"& á n"& d‹ œ Šn lim Ä_
" n‹
'
1
0
x"& dx
c1"& 2"& á n"& d
c1"& 2"& á n"& d‹ œ Šn lim n‹ Ä_
'
0
1
x"& dx œ _ (see part (b) above)
27. (a) Let the polygon be inscribed in a circle of radius r. If we draw a radius from the center of the circle (and the polygon) to each vertex of the polygon, we have n isosceles triangles formed (the equal sides are equal to r, the radius of the circle) and a vertex angle of )n where )n œ 2n1 . The area of each triangle is An œ
" # # r
sin )n Ê the area of the polygon is A œ nAn œ
(b) n lim A œ n lim Ä_ Ä_
nr# #
sin
21 n
œ n lim Ä_
n1r# 21
sin
21 n
nr# #
nr# 21 # sin n . sin ˆ 2n1 ‰ # ˆ 2n1 ‰ œ a1r b
sin )n œ
œ n lim a1 r # b Ä_
lim
2 1 În Ä 0
sin ˆ 2n1 ‰ ˆ 2n1 ‰
œ 1 r#
'x cos 2t dt " œ sin x ' x cos 2t dt " Ê yw œ cos x cosa2xb; when x œ 1 we have 1 yw œ cos 1 cosa21b œ " " œ #. And yww œ sin x 2sina2xb; when x œ 1, y œ sin 1 ' cos 2t dt " x
28. y œ sin x
1
1
œ ! ! " œ ". Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
357
358
Chapter 5 Integration
' ga$b œ '
29. (a) ga"b œ (b)
1
1
$
1
(c) ga"b œ
fatb dt œ ! fatb dt œ "# a#ba"b œ "
' fatb dt œ ' fatb dt œ "% a1 ## b œ 1 1
1
1
1
(d) gw axb œ faxb œ ! Ê x œ $, ", $ and the sign chart for gw axb œ faxb is relative maximum at x œ ". (e) gw a"b œ fa"b œ # is the slope and ga"b œ
± ± ± . So g has a 3 1 3
' fatb dt œ 1, by (c). Thus the equation is y 1 œ #ax "b 1
1
y œ #x # 1 . (f) gww axb œ f w axb œ ! at x œ " and gww axb œ f w axb is negative on a$ß "b and positive on a"ß "b so there is an inflection point for g at x œ ". We notice that gww axb œ f w axb ! for x on a"ß #b and gww axb œ f w axb ! for x on a#ß %b, even though gww a#b does not exist, g has a tangent line at x œ #, so there is an inflection point at x œ #. (g) g is continuous on Ò$ß %Ó and so it attains its absolute maximum and minimum values on this interval. We saw in (d) that gw axb œ ! Ê x œ $, ", $. We have that ga$b œ
' $ fatb dt œ '$" fatb dt œ 1## 1
#
œ #1
' fatb dt œ ! $ ga$b œ ' fatb dt œ " % ga%b œ ' fatb dt œ " "# † " † " œ "#
ga"b œ
1
1
1
1
Thus, the absolute minimum is #1 and the absolute maximum is !. Thus, the range is Ò#1ß !Ó.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 5 Additional and Advanced Exercises NOTES:
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
359
360
Chapter 5 Integration
NOTES:
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
CHAPTER 6 APPLICATIONS OF DEFINITE INTEGRALS 6.1 VOLUMES BY SLICING AND ROTATION ABOUT AN AXIS 1. (a) A œ 1(radius)# and radius œ È1 x# Ê A(x) œ 1 a1 x# b (b) A œ width † height, width œ height œ 2È1 x# Ê A(x) œ 4 a1 x# b (diagonal)# ; #
(c) A œ (side)# and diagonal œ È2(side) Ê A œ (d) A œ
È3 4
diagonal œ 2È1 x# Ê A(x) œ 2 a1 x# b
(side)# and side œ 2È1 x# Ê A(x) œ È3 a1 x# b
2. (a) A œ 1(radius)# and radius œ Èx Ê A(x) œ 1x (b) A œ width † height, width œ height œ 2Èx Ê A(x) œ 4x (diagonal)# ; #
(c) A œ (side)# and diagonal œ È2(side) Ê A œ (d) A œ
È3 4
(side)# and side œ 2Èx Ê A(x) œ È3x
(diagonal)# #
3. A(x) œ
diagonal œ 2Èx Ê A(x) œ 2x
œ
ˆ È x ˆ È x ‰ ‰ # #
œ 2x (see Exercise 1c); a œ 0, b œ 4;
V œ 'a A(x) dx œ '0 2x dx œ cx# d ! œ 16 b
4
1(diameter)# 4
4. A(x) œ
œ
%
1 c a2 x # b x # d 4
#
œ
1c2 a1 x# bd 4
#
œ 1 a1 2x# x% b ; a œ 1, b œ 1;
V œ 'a A(x) dx œ 'c1 1 a1 2x# x% b dx œ 1 ’x 23 x$ b
1
" x& 5 “ "
#
œ 21 ˆ1
2 3
5" ‰ œ
161 15
#
5. A(x) œ (edge)# œ ’È1 x# ŠÈ1 x# ‹“ œ Š2È1 x# ‹ œ 4 a1 x# b ; a œ 1, b œ 1; V œ 'a A(x) dx œ 'c1 4a1 x# b dx œ 4 ’x b
1
#
(diagonal)# #
6. A(x) œ
œ
œ
#
Š2È1 x# ‹
V œ 'a A(x) dx œ 2'c1 a1 x# b dx œ 2 ’x 1
" #
7. (a) STEP 1) A(x) œ
#
" x$ 3 “ "
(side) † (side) † ˆsin 13 ‰ œ
STEP 2) a œ 0, b œ 1
œ 8 ˆ1 "3 ‰ œ
16 3
#
’È1 x# ŠÈ1 x# ‹“
b
" x$ 3 “ "
" #
œ 2 a1 x# b (see Exercise 1c); a œ 1, b œ 1; œ 4 ˆ1 "3 ‰ œ
8 3
† Š2Èsin x‹ † Š2Èsin x‹ ˆsin 13 ‰ œ È3 sin x
STEP 3) V œ 'a A(x) dx œ È3 '0 sin x dx œ ’È3 cos x“ œ È3(1 1) œ 2È3 1
1
b
!
#
(b) STEP 1) A(x) œ (side) œ Š2Èsin x‹ Š2Èsin x‹ œ 4 sin x STEP 2) a œ 0, b œ 1
STEP 3) V œ 'a A(x) dx œ '0 4 sin x dx œ c4 cos xd 1! œ 8 1
b
#
8. (a) STEP 1) A(x) œ 1(diameter) œ 14 (sec x tan x)# œ 4 sin x ‘ œ 14 sec# x asec# x 1b 2 cos #x STEP 2) a œ 13 , b œ
asec# x tan# x 2 sec x tan xb
1 3
STEP 3) V œ 'a A(x) dx œ 'c1Î3 b
1 4
1Î3
1 4
ˆ2 sec# x 1
2 sin x ‰ cos# x
dx œ
1 4
2 tan x x 2 ˆ cos" x ‰‘1Î$ 1Î$
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
362
Chapter 6 Applications of Definite Integrals œ
1 4
’2È3
1 3
2 Š ˆ "" ‰ ‹ Š2È3 #
1 3
2 Š ˆ "" ‰ ‹‹“ œ #
(b) STEP 1) A(x) œ (edge)# œ (sec x tan x)# œ ˆ2 sec# x 1 2 STEP 2) a œ 13 , b œ
1 3
STEP 3) V œ 'a A(x) dx œ 'c1Î3 ˆ2 sec# x 1 1Î3
b
1 4
9. A(y) œ
(diameter)# œ
1 4
#
ŠÈ5y# 0‹ œ
c œ 0, d œ 2; V œ 'c A(y) dy œ '0 d
#
&
œ ’ˆ 541 ‰ Š y5 ‹“ œ !
" #
10. A(y) œ
2
1 4
51 4
51 4
Š4È3
21 3 ‹
sin x ‰ cos# x
dx œ 2 Š2È3 13 ‹ œ 4È3
21 3
y% ;
y% dy
a2& 0b œ 81
" #
(leg)(leg) œ
# È1 y# ˆÈ1 y# ‰‘ œ
V œ 'c A(y) dy œ 'c1 2a1 y# b dy œ 2 ’y d
2 sin x ‰ cos# x
1 4
1
" y$ 3 “ "
" #
#
ˆ2È1 y# ‰ œ 2 a1 y# b ; c œ 1, d œ 1;
œ 4 ˆ1 "3 ‰ œ
8 3
11. (a) It follows from Cavalieri's Principle that the volume of a column is the same as the volume of a right prism with a square base of side length s and altitude h. Thus, STEP 1) A(x) œ (side length)# œ s# ; STEP 2) a œ 0, b œ h; STEP 3) V œ 'a A(x) dx œ '0 s# dx œ s# h b
h
(b) From Cavalieri's Principle we conclude that the volume of the column is the same as the volume of the prism described above, regardless of the number of turns Ê V œ s# h 12. 1) The solid and the cone have the same altitude of 12. 2) The cross sections of the solid are disks of diameter x ˆ x# ‰ œ x# . If we place the vertex of the cone at the origin of the coordinate system and make its axis of symmetry coincide with the x-axis then the cone's cross sections will be circular disks of diameter x ˆ x‰ x 4 4 œ # (see accompanying figure). 3) The solid and the cone have equal altitudes and identical parallel cross sections. From Cavalieri's Principle we conclude that the solid and the cone have the same volume. 13. R(x) œ y œ 1 œ 1 ˆ2
4 2
14. R(y) œ x œ
3y #
x #
8 ‰ 12
# Ê V œ '0 1[R(x)]# dx œ 1'0 ˆ1 x# ‰ dx œ 1'0 Š1 x 2
œ
2
2
x# 4‹
dx œ 1 ’x
x# #
21 3
‰ dy œ 1' Ê V œ '0 1[R(y)]# dy œ 1'0 ˆ 3y # 0 2
15. R(x) œ tan ˆ 14 y‰ ; u œ
1 4
y Ê du œ
2
1 4
#
2
9 4
#
y# dy œ 1 34 y$ ‘ ! œ 1 †
3 4
dy Ê 4 du œ 1 dy; y œ 0 Ê u œ 0, y œ 1 Ê u œ
# x$ 12 “ !
† 8 œ 61 1 4
;
1Î% V œ '0 1[R(y)]# dy œ 1'0 tan ˆ 14 y‰‘ dy œ 4 '0 tan# u du œ 4 '0 a1 sec# ub du œ 4[u tan u]! 1
1
#
1Î4
1Î4
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 6.1 Volumes by Slicing and Rotation About an Axis œ 4 ˆ 14 1 0‰ œ 4 1 1 #
16. R(x) œ sin x cos x; R(x) œ 0 Ê a œ 0 and b œ œ 1'0
1Î2
xœ
1 #
(sin x cos x)# dx œ 1 '0
1Î2
Ê u œ 1‘ Ä V œ 1'0
1
" 8
(sin 2x)# 4
are the limits of integration; V œ '0
1Î2
dx; u œ 2x Ê du œ 2 dx Ê
sin# u du œ
1 8
#u
" 4
sin
1 2u‘ !
œ
1 8
ˆ 1#
du 8
œ
dx 4
1[R(x)]# dx
; x œ 0 Ê u œ 0,
0‰ 0‘ œ
1# 16
17. R(x) œ x# Ê V œ '0 1[R(x)]# dx œ 1 '0 ax# b dx 2
2
œ 1 '0 x% dx œ 1 ’ x5 “ œ 2
#
&
#
321 5
!
18. R(x) œ x$ Ê V œ '0 1[R(x)]# dx œ 1'0 ax$ b dx 2
2
œ 1 '0 x' dx œ 1 ’ x7 “ œ 2
(
# !
#
1281 7
19. R(x) œ È9 x# Ê V œ 'c3 1[R(x)]# dx œ 1 'c3 a9 x# b dx 3
$ x$ 3 “ $
œ 1 ’9x
3
œ 21 9(3)
27 ‘ 3
œ 2 † 1 † 18 œ 361
20. R(x) œ x x# Ê V œ '0 1[R(x)]# dx œ 1'0 ax x# b dx 1
1
œ 1'0 ax# 2x$ x% b dx œ 1 ’ x3 1
œ 1 ˆ 13
$
" #
5" ‰ œ
1 30
(10 15 6) œ
21. R(x) œ Ècos x Ê V œ '0
1Î2
1Î#
œ 1 csin xd !
2x% 4
1 30
#
" x& 5 “!
1[R(x)]# dx œ 1'0 cos x dx 1Î2
œ 1(1 0) œ 1
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363
364
Chapter 6 Applications of Definite Integrals 1Î4
1Î4
22. R(x) œ sec x Ê V œ 'c1Î4 1[R(x)]# dx œ 1 '1Î4 sec# x dx 1Î%
œ 1 ctan xd 1Î% œ 1[1 (1)] œ 21
23. R(x) œ È2 sec x tan x Ê V œ œ1
'01Î4 1[R(x)]# dx
'01Î4 ŠÈ2 sec x tan x‹# dx
œ 1 '0 Š2 2È2 sec x tan x sec# x tan# x‹ dx 1Î4
œ 1 Œ'0 2 dx 2È2 '0 sec x tan x dx 1Î4
1Î%
œ 1 Œ[2x]!
'01Î4 (tan x)# sec# x dx
1Î4
1Î%
2È2 [sec x]!
$
’ tan3 x “
1Î%
!
œ 1 ’ˆ 1# 0‰ 2È2 ŠÈ2 1‹ "3 a1$ 0b“ œ 1 Š 1# 2È2
11 3 ‹
24. R(x) œ 2 2 sin x œ 2(1 sin x) Ê V œ '0 1[R(x)]# dx 1Î2
œ 1 '0 4(1 sin x)# dx œ 41 '0 a1 sin# x 2 sin xb dx 1Î2
1Î2
œ 41'0 1 "# (1 cos 2x) 2 sin x‘ dx 1Î2
œ 41'0 ˆ 3# 1Î2
2 sin x‰
cos 2x 2
1Î# œ 41 3# x sin42x 2 cos x‘ ! œ 41 ˆ 341 0 0‰ (0 0 2)‘ œ 1(31 8)
25. R(y) œ È5 † y# Ê V œ 'c1 1[R(y)]# dy œ 1 'c1 5y% dy 1
1
"
œ 1 cy& d " œ 1[1 (1)] œ 21
26. R(y) œ y$Î# Ê V œ '0 1[R(y)]# dy œ 1'0 y$ dy 2
%
2
#
œ 1 ’ y4 “ œ 41 !
27. R(y) œ È2 sin 2y Ê V œ '0 1[R(y)]# dy 1Î2
œ 1'0 2 sin 2y dy œ 1 c cos 2yd ! 1Î2
1Î#
œ 1[1 (1)] œ 21
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 6.1 Volumes by Slicing and Rotation About an Axis 28. R(y) œ Écos
1y 4
Ê V œ 'c2 1[R(y)]# dy 0
œ 1 'c2 cos ˆ 14y ‰ dy œ 4 sin 0
29. R(y) œ
2 y1
1y ‘ ! 4 #
œ 4[0 (1)] œ 4
Ê V œ '0 1[R(y)]# dy œ 41 '0 3
3
" (y 1)#
dy
$
" ‘ œ 41 ’ y" 1 “ œ 41 4 (1) œ 31 !
30. R(y) œ
È2y y # 1
Ê V œ '0 1[R(y)]# dy œ 1'0 2y ay# 1b 1
1
#
dy;
#
cu œ y 1 Ê du œ 2y dy; y œ 0 Ê u œ 1, y œ 1 Ê u œ 2d Ä V œ 1'1 u# du œ 1 "u ‘ " œ 1 #" (1)‘ œ 2
#
1 #
31. For the sketch given, a œ 1# , b œ 1# ; R(x) œ 1, r(x) œ Ècos x; V œ 'a 1 a[R(x)]# [r(x)]# b dx b
œ 'c1Î2 1(1 cos x) dx œ 21'0 (1 cos x) dx œ 21[x sin x]! 1Î2
1Î2
1Î#
œ 21 ˆ 1# 1‰ œ 1# 21
32. For the sketch given, c œ 0, d œ 14 ; R(y) œ 1, r(y) œ tan y; V œ 'c 1 a[R(y)]# [r(y)]# b dy d
œ 1'0 a1 tan# yb dy œ 1 '0 a2 sec# yb dy œ 1[2y tan y]! 1Î4
1Î4
33. r(x) œ x and R(x) œ 1 Ê V œ œ '0 1 a1 x# b dx œ 1 ’x 1
1Î%
œ 1 ˆ 1# 1‰ œ
1# #
1
'01 1 a[R(x)]# [r(x)]# b dx
" x$ 3 “!
œ 1 ˆ1 "3 ‰ 0‘ œ
21 3
34. r(x) œ 2Èx and R(x) œ 2 Ê V œ '0 1 a[R(x)]# [r(x)]# b dx 1
œ 1'0 (4 4x) dx œ 41’x 1
" x# # “!
œ 41 ˆ1 "# ‰ œ 21
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365
366
Chapter 6 Applications of Definite Integrals
35. r(x) œ x# 1 and R(x) œ x 3
Ê V œ 'c1 1 a[R(x)]# [r(x)]# b dx 2
œ 1'c1 ’(x 3)# ax# 1b “ dx 2
#
œ 1 'c1 cax# 6x 9b ax% 2x# 1bd dx 2
œ 1 'c1 ax% x# 6x 8b dx 2
&
œ 1 ’ x5
x$ 3
œ 1 ˆ 32 5
8 3
#
6x# #
8x“
24 #
16‰ ˆ 5"
"
" 3
6 #
‰ ˆ 5†30533 ‰ œ 8‰‘ œ 1 ˆ 33 5 3 28 3 8 œ 1
36. r(x) œ 2 x and R(x) œ 4 x#
Ê V œ 'c1 1 a[R(x)]# [r(x)]# b dx 2
œ 1'c1 ’a4 x# b (2 x)# “ dx 2
#
œ 1 'c1 ca16 8x# x% b a4 4x x# bd dx 2
œ 1'c1 a12 4x 9x# x% b dx 2
œ 1 ’12x 2x# 3x$ œ 1 ˆ24 8 24
# x& 5 “ "
32 ‰ 5
ˆ12 2 3 "5 ‰‘ œ 1 ˆ15
33 ‰ 5
œ
1081 5
37. r(x) œ sec x and R(x) œ È2
Ê V œ 'c1Î4 1 a[R(x)]# [r(x)]# b dx 1Î4
œ 1 'c1Î4 a2 sec# xb dx œ 1[2x tan x]1Î% 1Î4
1Î%
œ 1 ˆ 1# 1‰ ˆ 1# 1‰‘ œ 1(1 2)
38. R(x) œ sec x and r(x) œ tan x
Ê V œ '0 1 a[R(x)]# [r(x)]# b dx 1
œ 1 '0 asec# x tan# xb dx œ 1 '0 1 dx œ 1[x]!" œ 1 1
1
39. r(y) œ 1 and R(y) œ 1 y
Ê V œ '0 1 a[R(y)]# [r(y)]# b dy 1
œ 1'0 c(1 y)# 1d dy œ 1 '0 a1 2y y# 1b dy 1
1
œ 1 '0 a2y y# b dy œ 1 ’y# 1
" y$ 3 “!
œ 1 ˆ1 3" ‰ œ
41 3
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
1171 5
Section 6.1 Volumes by Slicing and Rotation About an Axis 40. R(y) œ 1 and r(y) œ 1 y Ê V œ '0 1 a[R(y)]# [r(y)]# b dy 1
œ 1'0 c1 (1 y)# d dy œ 1'0 c1 a1 2y y# bd dy 1
1
œ 1'0 a2y y# b dy œ 1 ’y# 1
" y$ 3 “!
œ 1 ˆ1 "3 ‰ œ
21 3
41. R(y) œ 2 and r(y) œ Èy Ê V œ '0 1 a[R(y)]# [r(y)]# b dy 4
œ 1'0 (4 y) dy œ 1 ’4y 4
% y# 2 “!
œ 1(16 8) œ 81
42. R(y) œ È3 and r(y) œ È3 y#
È3
Ê V œ '0
È3
œ 1 '0
$
œ 1 ’ y3 “
1 a[R(y)]# [r(y)]# b dy
È3
c3 a3 y# bd dy œ 1'0 È$ !
y# dy
œ 1È3
43. R(y) œ 2 and r(y) œ 1 Èy Ê V œ '0 1 a[R(y)]# [r(y)]# b dy 1
œ 1'0 ’4 ˆ1 Èy‰ “ dy 1
#
œ 1 '0 ˆ4 1 2Èy y‰ dy 1
œ 1 '0 ˆ3 2Èy y‰ dy 1
œ 1 ’3y 43 y$Î# œ 1 ˆ3
" y# # “!
"# ‰ œ 1 ˆ 18683 ‰ œ
4 3
71 6
44. R(y) œ 2 y"Î$ and r(y) œ 1
Ê V œ '0 1 a[R(y)]# [r(y)]# b dy 1
# œ 1'0 ’ˆ2 y"Î$ ‰ 1“ dy 1
œ 1'0 ˆ4 4y"Î$ y#Î$ 1‰ dy 1
œ 1 '0 ˆ3 4y"Î$ y#Î$ ‰ dy 1
œ 1 ’3y 3y%Î$
" 3y&Î$ 5 “!
œ 1 ˆ3 3 53 ‰ œ
31 5
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367
368
Chapter 6 Applications of Definite Integrals
45. (a) r(x) œ Èx and R(x) œ 2 Ê V œ '0 1 a[R(x)]# [r(x)]# b dx 4
œ 1'0 (4 x) dx œ 1 ’4x 4
(b) r(y) œ 0 and R(y) œ y#
% x# # “!
œ 1(16 8) œ 81
Ê V œ '0 1 a[R(y)]# [r(y)]# b dy 2
œ 1'0 y% dy œ 1 ’ y5 “ œ 2
&
# !
321 5
# (c) r(x) œ 0 and R(x) œ 2 Èx Ê V œ '0 1 a[R(x)]# [r(x)]# b dx œ 1'0 ˆ2 Èx‰ dx 4
œ 1'0 ˆ4 4Èx x‰ dx œ 1 ’4x 4
4
8x$Î# 3
% x# # “!
œ 1 ˆ16
64 3
16 ‰ #
œ
81 3
(d) r(y) œ 4 y# and R(y) œ 4 Ê V œ '0 1 a[R(y)]# [r(y)]# b dy œ 1 '0 ’16 a4 y# b “ dy 2
2
œ 1 '0 a16 16 8y# y% b dy œ 1 '0 a8y# y% b dy œ 1 ’ 83 y$ 2
2
46. (a) r(y) œ 0 and R(y) œ 1
# y& 5 “!
#
œ 1 ˆ 64 3
32 ‰ 5
œ
2241 15
y #
Ê V œ '0 1 a[R(y)]# [r(y)]# b dy 2
# œ 1'0 ˆ1 y# ‰ dy œ 1'0 Š1 y 2
œ 1 ’y
2
y# #
# y$ 12 “ !
œ 1 ˆ#
(b) r(y) œ 1 and R(y) œ 2
4 2
8 ‰ 12
y# 4‹
œ
dy
21 3
y #
# Ê V œ '0 1 a[R(y)]# [r(y)]# b dy œ 1 '0 ’ˆ2 y# ‰ 1“ dy œ 1 '0 Š4 2y 2
2
œ 1'0 Š3 2y 2
y# 4‹
dy œ 1 ’3y y#
# y$ 12 “ !
2
œ 1 ˆ6 4
8 ‰ 12
œ 1 ˆ2 23 ‰ œ
y# 4
1‹ dy
81 3
47. (a) r(x) œ 0 and R(x) œ 1 x#
Ê V œ 'c1 1 a[R(x)]# [r(x)]# b dx 1
œ 1 'c1 a1 x# b dx œ 1 'c1 a1 2x# x% b dx 1
œ 1 ’x
1
#
2x$ 3
" x& 5 “ "
103 ‰ œ 21 ˆ 1515 œ
œ 21 ˆ1
2 3
15 ‰
161 15
(b) r(x) œ 1 and R(x) œ 2 x# Ê V œ 'c1 1 a[R(x)]# [r(x)]# b dx œ 1 'c1 ’a2 x# b 1“ dx 1
1
œ 1 'c1 a4 4x# x% 1b dx œ 1'c1 a3 4x# x% b dx œ 1 ’3x 43 x$ 1
œ
21 15
1
(45 20 3) œ
561 15
#
" x& 5 “ "
œ 21 ˆ3
4 3
15 ‰
2 3
15 ‰
(c) r(x) œ 1 x# and R(x) œ 2 Ê V œ 'c1 1 a[R(x)]# [r(x)]# b dx œ 1 'c1 ’4 a1 x# b “ dx 1
1
œ 1 'c1 a4 1 2x# x% b dx œ 1'c1 a3 2x# x% b dx œ 1 ’3x 23 x$ 1
œ
21 15
1
(45 10 3) œ
641 15
#
" x& 5 “ "
œ 21 ˆ3
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Section 6.1 Volumes by Slicing and Rotation About an Axis
369
48. (a) r(x) œ 0 and R(x) œ hb x h
Ê V œ '0 1 a[R(x)]# [r(x)]# b dx b
# œ 1 '0 ˆ hb x h‰ dx b
œ 1'0 Š hb# x# b
#
$
x œ 1h# ’ 3b #
x# b
2h# b
x h# ‹ dx b
x“ œ 1h# ˆ b3 b b‰ œ !
1 h# b 3
# (b) r(y) œ 0 and R(y) œ b ˆ1 yh ‰ Ê V œ '0 1 a[R(y)]# [r(y)]# b dy œ 1b# '0 ˆ1 yh ‰ dy h
œ 1b# '0 Š1 h
2y h
y# h# ‹
dy œ 1b# ’y
y# h
h
h
y$ 3h# “ !
1 b# h 3
œ 1b# ˆh h 3h ‰ œ
49. R(y) œ b Èa# y# and r(y) œ b Èa# y# Ê V œ 'ca 1 a[R(y)]# [r(y)]# b dy a
œ 1 'ca ’ˆb Èa# y# ‰ ˆb Èa# y# ‰ “ dy #
a
#
œ 1 'ca 4bÈa# y# dy œ 4b1'ca Èa# y# dy a
a
1a# #
œ 4b1 † area of semicircle of radius a œ 4b1 †
œ 2a# b1#
50. (a) A cross section has radius r œ È#y and area 1r# œ #1y. The volume is '0 #1ydy œ 1 cy# d ! œ #&1. &
(b) Vahb œ ' Aahbdh, so
dV dh
œ Aahb. Therefore
For h œ %, the area is #1a%b œ )1, so
dh dt
œ
dV dt
" )1
œ
dV dh
†
œ Aahb †
dh dt
$
$ )1
† $ units sec œ
hca
51. (a) R(y) œ Èa# y# Ê V œ 1'ca aa# y# b dy œ 1 ’a# y œ 1 ’a# h "3 ah$ 3h# a 3ha# a$ b dV $ dt œ 0.2 m /sec dV # dh œ 101h 1h
(b) Given Ê
and a œ 5 m, find Ê
dV dt
œ
dV dh
†
dh dt
a$ 3“
œ 1 Ša# h
†
so
dh dt
œ
" A ah b
†
dV dt .
units$ sec . hca
y$ 3 “ ca
h$ 3
dh dt ,
&
œ 1 ’a# h a$
h# a ha# ‹ œ
(h a)$ 3
Ša$
a$ 3 ‹“
1h# (3a h) 3
#
From part (a), V(h) œ 1h (153 h) œ 51h# 13h dh ¸ 0.2 " " œ 1h(10 h) dh dt Ê dt hœ4 œ 41(10 4) œ (201)(6) œ 1#01 m/sec. dh ¸ dt hœ4 .
$
52. Suppose the solid is produced by revolving y œ 2 x about the y-axis. Cast a shadow of the solid on a plane parallel to the xy-plane. Use an approximation such as the Trapezoid Rule, to #
estimate 'a 1cRaybd# dy ¸ ! 1Œ #k ˜y. b
n
d^
kœ"
53. The cross section of a solid right circular cylinder with a cone removed is a disk with radius R from which a disk of radius h has been removed. Thus its area is A" œ 1R# 1h# œ 1 aR# h# b . The cross section of the hemisphere is a disk of #
radius ÈR# h# . Therefore its area is A# œ 1 ŠÈR# h# ‹ œ 1 aR# h# b . We can see that A" œ A# . The altitudes of both solids are R. Applying Cavalieri's Principle we find Volume of Hemisphere œ (Volume of Cylinder) (Volume of Cone) œ a1R# b R "3 1 aR# b R œ
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
2 3
1 R$ .
370
Chapter 6 Applications of Definite Integrals
54. R(x) œ
rx h
Ê V œ '0 1[R(x)]# dx œ 1'0 h
h # # r x
dx œ
h#
1r# h#
h
$
#
$
’ x3 “ œ Š 1hr# ‹ Š h3 ‹ œ !
" 3
1r# h, the volume of
a cone of radius r and height h. c7
c7
55. R(y) œ È256 y# Ê V œ 'c16 1[R(y)]# dy œ 1'c16 a256 y# b dy œ 1 ’256y œ 1 ’(256)(7) 56. R(x) œ œ
1 144
x 1#
7$ 3
Š(256)(16)
16$ 3 ‹“
$
È36 x# Ê V œ ' 1[R(x)]# dx œ 1' 0 0 6
' x& 5 “!
’12x$
œ
1 144
6
Š12 † 6$
6& 5‹
œ
16$ 3 ‹
œ 1 Š 73 256(16 7)
1 †6 $ 144
x# 144
ˆ12
a36 x# b dx œ
36 ‰ 5
1 144
( y$ 3 “ "'
œ 10531 cm$ ¸ 3308 cm$
'06 a36x# x% b dx
1 ‰ ˆ 6036 ‰ œ ˆ 196 œ 144 5
361 5
cm$ . The plumb bob will
weigh about W œ (8.5) ˆ 3651 ‰ ¸ 192 gm, to the nearest gram. 57. (a) R(x) œ kc sin xk , so V œ 1'0 [R(x)]# dx œ 1'0 (c sin x)# dx œ 1'0 ac# 2c sin x sin# xb dx 1
1
œ 1'0 ˆc# 2c sin x
'1
1
(b)
1
1cos 2x ‰ dx œ 1 0 ˆc# "# 2c sin x cos#2x ‰ dx # 1 œ 1 ˆc# "# ‰ x 2c cos x sin42x ‘ ! œ 1 ˆc# 1 1# 2c 0‰ (0 2c 0)‘ œ 1 ˆc# 1 1# 4c‰ . Let 2 V(c) œ 1 ˆc# 1 1# 4c‰ . We find the extreme values of V(c): dV dc œ 1(2c1 4) œ 0 Ê c œ 1 is a critical # # point, and V ˆ 12 ‰ œ 1 ˆ 14 1# 18 ‰ œ 1 ˆ 1# 14 ‰ œ 1# 4; Evaluate V at the endpoints: V(0) œ 1# and # # V(1) œ 1 ˆ 3# 1 4‰ œ 1# (4 1)1. Now we see that the function's absolute minimum value is 1# 4, taken on at the critical point c œ 12 . (See also the accompanying graph.) # From the discussion in part (a) we conclude that the function's absolute maximum value is 1# , taken on at
the endpoint c œ 0. (c) The graph of the solid's volume as a function of c for 0 Ÿ c Ÿ 1 is given at the right. As c moves away from [!ß "] the volume of the solid increases without bound. If we approximate the solid as a set of solid disks, we can see that the radius of a typical disk increases without bounds as c moves away from [0ß 1]. 58. (a) R(x) œ 1
œ 1 'c4 Š1 4
œ 1 ’x
x$ 24
œ 21 ˆ4
Ê V œ 'c4 1[R(x)]# dx 4
x# 16
8 3
x# 16 ‹
#
dx œ 1'c4 Š1 4
%
x& 5†16# “ %
45 ‰ œ
21 15
œ 21 Š4
x# 8
x% 16# ‹
4$ 24
4& 5†16# ‹
(60 40 12) œ
641 15
dx
ft$
(b) The helicopter will be able to fly ˆ 64151 ‰ (7.481)(2) ¸ 201 additional miles. 6.2 VOLUME BY CYLINDRICAL SHELLS 1. For the sketch given, a œ 0, b œ 2;
shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x Š1 b
2
x# 4‹
dx œ 21'0 Šx
x# 4‹
dx œ 21'0 Š2x
2
x$ 4‹
œ 21 † 3 œ 61 2. For the sketch given, a œ 0, b œ 2;
shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x Š2 b
2
2
#
dx œ 21 ’ x#
x$ 4‹
# x% 16 “ !
dx œ 21 ’x#
œ 21 ˆ 4#
# x% 16 “ !
16 ‰ 16
œ 21(4 1) œ 61
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 6.2 Volume by Cylindrical Shells 3. For the sketch given, c œ 0, d œ È2;
È2
shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 d
È2
21y † ay# b dy œ 21'0
%
y$ dy œ 21 ’ y4 “
4. For the sketch given, c œ 0, d œ È3;
È3
È3
shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21y † c3 a3 y# bd dy œ 21 '0 d
È# !
œ 21
%
y$ dy œ 21 ’ y4 “
È3 !
œ
5. For the sketch given, a œ 0, b œ È3;
È3
shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x † ŠÈx# 1‹ dx; b
’u œ x# 1 Ê du œ 2x dx; x œ 0 Ê u œ 1, x œ È3 Ê u œ 4“ Ä V œ 1'1 u"Î# du œ 1 23 u$Î# ‘ " œ %
4
21 3
ˆ4$Î# 1‰ œ ˆ 231 ‰ (8 1) œ
141 3
6. For the sketch given, a œ 0, b œ 3;
shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x Š Èx9x ‹ dx; $9 b
3
cu œ x$ 9 Ê du œ 3x# dx Ê 3 du œ 9x# dx; x œ 0 Ê u œ 9, x œ 3 Ê u œ 36d Ä V œ 21 '9 3u"Î# du œ 61 2u"Î# ‘ * œ 121 ŠÈ36 È9‹ œ 361 $'
36
7. a œ 0, b œ 2;
shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x x ˆ x2 ‰‘ dx b
2
œ '0 21x# † 2
3 #
dx œ 1 '0 3x# dx œ 1 cx$ d ! œ 81 2
#
8. a œ 0, b œ 1;
shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x ˆ2x x2 ‰ dx b
1
œ 1 '0 2 Š 3x# ‹ dx œ 1 ' 3x# dx œ 1 cx$ d ! œ 1 1
1
#
"
0
9. a œ 0, b œ 1;
shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x c(2 x) x# d dx b
1
œ 21'0 a2x x# x$ b dx œ 21 ’x# 1
œ 21 ˆ1
" 3
4" ‰ œ 21 ˆ 12 124 3 ‰ œ
x$ 3
101 12
œ
" x% 4 “!
51 6
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
91 #
371
372
Chapter 6 Applications of Definite Integrals
10. a œ 0, b œ 1;
shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x ca2 x# b x# d dx b
1
œ 21'0 x a2 2x# b dx œ 41'0 ax x$ b dx 1
1
" x% 4 “!
#
œ 41 ’ x#
œ 41 ˆ "2 4" ‰ œ 1
11. a œ 0, b œ 1;
shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x Èx (2x 1)‘ dx b
1
" œ 21'0 ˆx$Î# 2x# x‰ dx œ 21 25 x&Î# 23 x$ "# x# ‘ ! 1
œ 21 ˆ 25
2 3
15 ‰ "# ‰ œ 21 ˆ 12 20 œ 30
71 15
12. a œ ", b œ 4;
shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '1 21x ˆ 32 x"Î# ‰ dx b
4
œ 31'1 x"Î# dx œ 31 23 x$Î# ‘ " œ 21 ˆ4$Î# "‰ %
4
œ 21(8 1) œ 141
13. (a) xf(x) œ œ
xf(x) œ œ
x†
sin x, 0 x Ÿ 1 0xŸ1 Ê xf(x) œ œ ; since sin 0 œ 0 we have 0, x œ 0 x, x œ 0
sin x x ,
sin x, 0 x Ÿ 1 Ê xf(x) œ sin x, 0 Ÿ x Ÿ 1 sin x, x œ 0
shell ‰ shell (b) V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x † f(x) dx and x † f(x) œ sin x, 0 Ÿ x Ÿ 1 by part (a) 1
b
Ê V œ 21'0 sin x dx œ 21[ cos x]1! œ 21( cos 1 cos 0) œ 41 1
tan# x x ,
tan# x, 0 x Ÿ 1/4 0 x Ÿ 14 Ê xg(x) œ œ ; since tan 0 œ 0 we have 0, x œ 0 x † 0, x œ 0 tan# x, 0 x Ÿ 1/4 Ê xg(x) œ tan# x, 0 Ÿ x Ÿ 1/4 xg(x) œ œ tan# x, x œ 0
14. (a) xg(x) œ œ
x†
shell ‰ shell (b) V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x † g(x) dx and x † g(x) œ tan# x, 0 Ÿ x Ÿ 1/4 by part (a) 1Î4
b
Ê V œ 21'0 tan# x dx œ 21'0 asec# x 1b dx œ 21[tan x x]! 1Î4
1Î4
1Î%
œ 21 ˆ1 14 ‰ œ
41 1 # #
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 6.2 Volume by Cylindrical Shells 15. c œ 0, d œ 2;
shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21y Èy (y)‘ dy d
2
œ 21'0 ˆy$Î# y# ‰ dy œ 21 ’ 2y5 2
&Î#
&
œ 21 ” 25 ŠÈ2‹ œ
161 15
2$ 3•
# y$ 3 “!
È
œ 21 Š 8 5 2 83 ‹ œ 161 Š
È2 5
"3 ‹
Š3È2 5‹
16. c œ 0, d œ 2;
shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21y cy# (y)ddy d
2
œ 21'0 ay$ y# b dy œ 21 ’ y4 2
%
œ 161 ˆ 56 ‰ œ
401 3
# y$ 3 “!
œ 161 ˆ 24 "3 ‰
17. c œ 0, d œ 2;
shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21y a2y y# bdy d
2
œ 21'0 a2y# y$ b dy œ 21 ’ 2y3 2
$
œ 321 ˆ "3 4" ‰ œ
321 12
œ
81 3
# y% 4 “!
œ 21 ˆ 16 3
"6 ‰ 4
18. c œ 0, d œ 1;
shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21y a2y y# ybdy d
1
œ 21'0 y ay y# b dy œ 21'0 ay# y$ b dy 1
1
$
œ 21 ’ y3
" y% “ 4 !
œ 21 ˆ 13 "4 ‰ œ
1 6
19. c œ 0, d œ 1;
shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ 21'0 y[y (y)]dy d
1
œ 21'0 2y# dy œ 1
41 3
"
cy$ d ! œ
41 3
20. c œ 0, d œ 2;
shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21 yˆy y2 ‰dy d
2
œ 21 '0
2
y2 2 dy
1
œ 13 c y3 d ! œ
81 3
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
373
374
Chapter 6 Applications of Definite Integrals
21. c œ 0, d œ 2;
shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21y c(2 y) y# d dy d
2
œ 21 '0 a2y y# y$ b dy œ 21 ’y# 2
œ 21 ˆ4
8 3
16 ‰ 4
1 6
œ
y$ 3
(48 32 48) œ
# y% 4 “!
161 3
22. c œ 0, d œ 1;
shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21y c(2 y) y# d dy d
1
œ 21'0 a2y y# y$ b dy œ 21 ’y# 1
œ 21 ˆ1
14 ‰ œ
1 3
1 6
(12 4 3) œ
y$ 3
51 6
" y% 4 “!
shell ‰ shell 23. (a) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21y † 12 ay# y$ b dy œ 241 '0 ay$ y% b dy œ 241 ’ y4 d
1
œ 241 ˆ 14 15 ‰ œ
241 20
œ
1
" y& 5 “!
%
61 5
shell ‰ shell (b) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21(1 y) c12 ay# y$ bd dy œ 241'0 (1 y) ay# y$ b dy d
1
1
œ 241'0 ay# 2y$ y% b dy œ 241 ’ y3 1
$
y% 2
" y& 5 “!
œ 241 ˆ "3
1 2
" ‰ 51 ‰ œ 241 ˆ 30 œ
41 5
shell ‰ shell (c) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21 ˆ 85 y‰ c12 ay# y$ bd dy œ 241 '0 ˆ 85 y‰ ay# y$ b dy d
1
œ 241'0 ˆ 85 y# 1
œ
241 12
13 5
1
8 $ y$ y% ‰ dy œ 241 ’ 15 y
13 20
y%
" y& 5 “!
8 œ 241 ˆ 15
13 20
241 60
15 ‰ œ
(32 39 12)
œ 21
shell ‰ shell (d) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21 ˆy 25 ‰ c12 ay# y$ bd dy œ 241'0 ˆy 25 ‰ ay# y$ b dy d
1
1
2 $ œ 241'0 ˆy$ y% 25 y# 25 y$ ‰ dy œ 241'0 ˆ 25 y# 35 y$ y% ‰ dy œ 241 ’ 15 y 1
1
2 œ 241 ˆ 15
3 20
15 ‰ œ
241 60
(8 9 12) œ
241 12
2
%
œ 21 ’ y4
#
y' 24 “ !
%
2' 24 ‹
œ 21 Š 24
#
%
œ 321 ˆ 4"
4 ‰ 24
dy œ '0 21y Šy# 2
y# # ‹“
2
œ 21 '0 Š2y# 2
y% 2
y$
y& 4‹
#
$
dy œ 21 ’ 2y3
y% 4
2
œ 21'0 Š5y# 54 y% y$ 2
y& 4‹
#
$
dy œ 21 ’ 5y3
y% 4
# y' #4 “ !
œ 21 ˆ 16 3
œ
2
œ 21'0 Šy$ 2
y& 4
58 y#
5 32
#
%
y% ‹ dy œ 21 ’ y4
%
y' #4
5y$ #4
2
y# # ‹“
16 4
64 ‰ 24
2
y' #4 “ ! y# # ‹“
32 10
dy œ '0 21(5 y) Šy# #
shell ‰ shell (d) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21 ˆy 58 ‰ ’ y# Š y4 d
81 3 y% 4‹
%
5y& 20
2
dy œ '0 21(2 y) Šy#
shell ‰ shell (c) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21(5 y) ’ y# Š y4 d
dy œ 21'0 Šy$
y# # ‹“
%
y& 10
y% 4‹
2 ‰ œ 321 ˆ 4" 6" ‰ œ 321 ˆ 24 œ
shell ‰ shell (b) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21(2 y) ’ y# Š y4 d
" y& 5 “!
y%
œ 21
shell ‰ shell 24. (a) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21y ’ y# Š y4 d
3 20
œ 21 ˆ 40 3
160 20
16 4
dy œ '0 21 ˆy 58 ‰ Šy#
# 5y& 160 “ !
2
œ 21 ˆ 16 4
64 24
40 24
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
81 5
y% 4‹
64 ‰ 24
dy
dy
œ 81 y% 4‹
160 ‰ 160
dy
œ 41
y& 4‹
dy
Section 6.2 Volume by Cylindrical Shells
375
shell ‰ shell 25. (a) About x-axis: V œ 'c 21 ˆ radius Š height ‹dy d
œ '0 21yˆÈy y‰dy œ 21'0 ˆy$Î# y# ‰dy 1
1
" œ 21 #& y&Î# "$ y$ ‘ ! œ 21ˆ #& "$ ‰ œ
#1 "&
shell ‰ shell About y-axis: V œ 'a 21 ˆ radius Š height ‹dx b
œ '0 21xax x# bdx œ 21'0 ax2 x3 bdx 1
1
$
œ 21’ x$
" x% % “!
œ 21ˆ "$ "% ‰ œ
1 '
(b) About x-axis: Raxb œ x and raxb œ x# Ê V œ 'a 1Raxb# raxb# ‘dx œ '0 1cx# x% ddx b
$
œ 1’ x$
" x& & “!
œ 1ˆ "$ "& ‰ œ
1
#1 "&
About y-axis: Rayb œ Èy and rayb œ y Ê V œ 'c 1Rayb# rayb# ‘dy œ '0 1cy y2 ddy d
#
œ 1’ y#
" y$ $ “!
œ 1ˆ "# "$ ‰ œ
1
1 '
# 26. (a) V œ 'a 1Raxb# raxb# ‘dx œ 1'0 ’ˆ #x #‰ x# “dx %
b
œ 1'0 ˆ $% x# #x %‰dx œ 1’ x% x# %x“ %
$
œ 1a"' "' "'b œ "'1
% !
shell ‰ shell (b) V œ 'a 21 ˆ radius Š height ‹dx œ '0 #1xˆ x# # x‰dx %
b
œ '0 #1xˆ# x# ‰dx œ #1'0 Š#x %
%
œ #1’x#
% x$ ' “!
'% ‰ '
œ #1ˆ"'
œ
x# # ‹dx $#1 $
shell ‰ shell (c) V œ 'a 21 ˆ radius Š height ‹dx œ '0 #1a% xbˆ x# # x‰dx œ '0 #1a% xbˆ# x# ‰dx œ #1'0 Š) %x %
b
œ #1’)x #x#
% x$ “ ' !
œ #1ˆ$# $#
%
'% ‰ '
%
x# # ‹dx
'%1 $
œ
# (d) V œ 'a 1Raxb# raxb# ‘dx œ 1'0 ’a) xb# ˆ' #x ‰ “dx œ 1'0 ’a'% "'x x# b Š$' 'x x% ‹“dx %
b
%
#
1'0 ˆ $% x# "!x #)‰dx œ 1’ x% &x# #)x“ œ 1"' a&ba"'b a(ba"'b‘ œ 1a$ba"'b œ %)1 %
%
$
!
shell ‰ shell 27. (a) V œ 'c 21 ˆ radius Š height ‹ dy œ '1 21y(y 1) dy d
2
œ 21'1 ay# yb dy œ 21 ’ y3 2
$
# y# # “"
œ 21 ˆ 83 42 ‰ ˆ "3 #" ‰‘ œ 21 ˆ 73 2 "# ‰ œ 13 (14 12 3) œ
shell ‰ shell (b) V œ 'a 21 ˆ radius Š height ‹ dx
51 3
b
œ '1 21x(2 x) dx œ 21'1 a2x x# b dx œ 21 ’x# 2
2
œ 21 ˆ 12 3 8 ‰ ˆ 3 3 " ‰‘ œ 21 ˆ 34 32 ‰ œ
41 3
# x$ 3 “"
œ 21 ˆ4 83 ‰ ˆ1 "3 ‰‘
shell ‰ shell ' ˆ 203 ‰ (c) V œ 'a 21 ˆ radius Š height ‹ dx œ '1 21 ˆ 10 3 x (2 x) dx œ 21 1 b
2
#
" $‘ 8 # ˆ 40 œ 21 20 3 x 3 x 3 x " œ 21 3
2
32 3
38 ‰ ˆ 20 3
8 3
16 3
x x# ‰ dx
3" ‰‘ œ 21 ˆ 33 ‰ œ 21
shell ‰ shell (d) V œ 'c 21 ˆ radius Š height ‹ dy œ '1 21(y 1)(y 1) dy œ 21'1 (y 1)# œ 21 ’ (y31) “ œ d
2
2
$
# "
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
21 3
376
Chapter 6 Applications of Definite Integrals
shell ‰ shell 28. (a) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21yay# 0b dy d
2
œ 21'0 y$ dy œ 21 ’ y4 “ œ 21 Š 24 ‹ œ 81 2
%
#
%
!
shell ‰ shell (b) V œ 'a 21 ˆ radius Š height ‹ dx b
œ '0 21x ˆ2 Èx‰ dx œ 21'0 ˆ2x x$Î# ‰ dx 4
4
%
2 †2 & 5 ‹
œ 21 x# 25 x&Î# ‘ ! œ 21 Š16 œ 21 ˆ16
64 ‰ 5
21 5
œ
321 5
(80 64) œ
shell ‰ shell (c) V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21(4 x) ˆ2 Èx‰ dx œ 21'0 ˆ8 4x"Î# 2x x$Î# ‰ dx b
4
4
%
œ 21 8x 83 x$Î# x# 25 x&Î# ‘ ! œ 21 ˆ32
64 3
16
64 ‰ 5
œ
21 15
(240 320 192) œ
21 15
shell ‰ shell (d) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21(2 y) ay# b dy œ 21 '0 a2y# y$ b dy œ 21 ’ 23 y$ d
2
œ 21 ˆ 16 3
16 ‰ 4
œ
321 12
2
81 3
(4 3) œ
(112) œ
2241 15
# y% 4 “!
shell ‰ shell 29. (a) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21yay y$ b dy d
1
œ '0 21 ay# y% b dy œ 21 ’ y3 1
œ
$
41 15
" y& “ 5 !
œ 21 ˆ "3 "5 ‰
shell ‰ shell (b) V œ 'c 21 ˆ radius Š height ‹ dy d
œ '0 21(1 y) ay y$ b dy 1
œ 21 '0 ay y# y$ y% b dy œ 21 ’ y# 1
#
y$ 3
y% 4
" y& 5 “!
œ 21 ˆ "#
" 3
" 4
5" ‰ œ
21 60
(30 20 15 12) œ
71 30
shell ‰ shell 30. (a) V œ 'c 21 ˆ radius Š height ‹dy d
œ '0 21y c1 ay y$ bddy 1
œ 21 '0 ay y# y% b dy œ 21 ’ y# 1
#
œ 21 ˆ "# œ
" 3
5" ‰ œ
21 30
y$ 3
" y& 5 “!
(15 10 6)
111 15
(b) Use the washer method: V œ 'c 1 cR# (y) r# (y)d dy œ '0 1 ’1# ay y$ b “ dy œ 1 '0 a1 y# y' 2y% b dy œ 1 ’y d
1
œ 1 ˆ1
" 3
" 7
25 ‰ œ
1 105
1
#
(105 35 15 42) œ
y$ 3
y( 7
971 105
" 2y& 5 “!
(c) Use the washer method: V œ 'c 1 cR# (y) r# (y)d dy œ '0 1 ’c1 ay y$ bd 0“ dy œ 1'0 ’1 2 ay y$ b ay y$ b “ dy d
1
œ 1'0 a1 y# y' 2y 2y$ 2y% b dy œ 1 ’y 1
œ
1 210
(70 30 105 2 † 42) œ
1
#
y$ 3
y( 7
y#
#
y% #
1211 210
" 2y& 5 “!
œ 1 ˆ1
" 3
" 7
1
" #
25 ‰
shell ‰ shell (d) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21(1 y) c1 ay y$ bd dy œ 21 '0 (1 y) a1 y y$ b dy d
1
1
œ 21'0 a1 y y$ y y# y% b dy œ 21'0 a1 2y y# y$ y% b dy œ 21 ’y y# 1
œ 21 ˆ1 1
1
" 3
" 4
5" ‰ œ
21 60
(20 15 12) œ
231 30
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
y$ 3
y% 4
" y& 5 “!
Section 6.2 Volume by Cylindrical Shells shell ‰ shell 31. (a) V œ 'c 21 ˆ radius Š height ‹dy œ '0 21y ˆÈ8y y# ‰ dy d
2
œ 21'0 Š2È2 y$Î# y$ ‹ dy œ 21 ’ 4 5 2 y&Î# È
2
# y% 4 “!
&
œ 21
4È2†ŠÈ2‹
2% 4
5
œ 21 † 4 ˆ 85 1‰ œ
81 5
$
œ 21 Š 4†52
(8 5) œ
4 †4 4 ‹
241 5
shell ‰ shell (b) V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x ŠÈx b
4
&
œ 21 Š 2†52
4% 3# ‹
'
œ 21 Š 25
2) 32 ‹
œ
1†2( 160
x# 8‹
dx œ 21'0 Šx$Î#
(32 20) œ
4
1†2* †3 160
œ
1†2% †3 5
œ
x$ 8‹
dx œ 21 ’ 25 x&Î#
481 5
shell ‰ shell 32. (a) V œ 'a 21 ˆ radius Š height ‹ dx b
œ '0 21x ca2x x# b xd dx 1
œ 21 '0 x ax x# b dx œ 21'0 ax# x$ b dx 1
$
1
œ 21 ’ x3
" x% 4 “!
œ 21 ˆ "3 4" ‰ œ
1 6
shell ‰ shell (b) V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21(1 x) ca2x x# b xd dx œ 21'0 (1 x) ax x# b dx b
1
1
œ 21 '0 ax 2x# x$ b dx œ 21 ’ x2 23 x$ 1
" x% 4 “!
#
œ 21 ˆ 12
2 3
"4 ‰ œ
21 1#
(6 8 3) œ
1 6
33. (a) V œ 'a 1 cR# (x) r# (x)d dx œ 1 '1Î16 ˆx"Î# 1‰ dx b
1
"
œ 1 2x"Î# x‘"Î"' œ 1 (2 1) ˆ2 † œ 1 ˆ1
7 ‰ 16
œ
" 4
" ‰‘ 16
91 16
shell ‰ shell (b) V œ 'a 21 ˆ radius Š height ‹ dy œ '1 21y Š y"% b
2
œ 21'1 ˆy$ 2
y ‰ 16
dy œ 21 ’ 12 y#
œ 21 ˆ "8 8" ‰ ˆ #" œ
21 32
(8 1) œ
91 16
" ‰‘ 3#
d
2
œ 1 "3 y$ œ
1 48
y ‘# 16 "
y# 32 “ "
" ‰ 32
" 16 ‹
dy
" œ 1 ˆ 24 8" ‰ ˆ 3"
(2 6 16 3) œ
dy
#
œ 21 ˆ 4"
34. (a) V œ 'c 1 cR# (y) r# (y)d dy œ '1 1 Š y"%
" 16 ‹
" ‰‘ 16
111 48
shell ‰ shell (b) V œ 'a 21 ˆ radius Š height ‹ dx œ '1Î4 21x Š È"x "‹ dx b
1
œ 21 '1Î4 ˆx"Î# x‰ dx œ 21 ’ 23 x$Î# 1
œ 21 ˆ 23 "# ‰ ˆ 23 †
" 8
" ‰‘ 3#
" x# 2 “ "Î%
œ 1 ˆ 43 1
" 6
" ‰ 16
œ
1 48
(4 † 16 48 8 3) œ
111 48
35. (a) H3=k: V œ V" V#
V" œ 'a 1[R" (x)]# dx and V# œ 'a 1[R# (x)]# with R" (x) œ É x 3 2 and R# (x) œ Èx, b"
b#
"
#
a" œ 2, b" œ 1; a# œ 0, b# œ 1 Ê two integrals are required (b) [ +=2/ sqrt(1-x^2);a := -1; b := 1; N := [2, 4, 8 ]; for n in N do xx := [seq( a+i*(b-a)/n, i=0..n )]; pts := [seq([x,f(x)],x=xx)]; L := simplify(add( distance(pts[i+1],pts[i]), i=1..n )); T := sprintf("#31(a) (Section 6.3)\nn=%3d L=%8.5f\n", n, L ); P[n] := plot( [f(x),pts], x=a..b, title=T ): end do: display( [seq(P[n],n=N)], insequence=true, scaling=constrained ); L := ArcLength( f(x), x=a..b, output=integral ): L = evalf( L );
# (b) # (a)
# (c)
37-40. Example CAS commands: Maple: with( plots ); with( student ); x := t -> t^3/3; y := t -> t^2/2; a := 0; b := 1; N := [2, 4, 8 ]; for n in N do tt := [seq( a+i*(b-a)/n, i=0..n )]; pts := [seq([x(t),y(t)],t=tt)]; Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
383
384
Chapter 6 Applications of Definite Integrals L := simplify(add( student[distance](pts[i+1],pts[i]), i=1..n )); T := sprintf("#37(a) (Section 6.3)\nn=%3d L=%8.5f\n", n, L ); P[n] := plot( [[x(t),y(t),t=a..b],pts], title=T ): end do: display( [seq(P[n],n=N)], insequence=true ); ds := t ->sqrt( simplify(D(x)(t)^2 + D(y)(t)^2) ): L := Int( ds(t), t=a..b ): L = evalf(L);
# (b) # (a)
# (c)
31-40. Example CAS commands: Mathematica: (assigned function and values for a, b, and n may vary) Clear[x, f] {a, b} = {1, 1}; f[x_] = Sqrt[1 x2 ] p1 = Plot[f[x], {x, a, b}] n = 8; pts = Table[{xn, f[xn]}, {xn, a, b, (b a)/n}]/ / N Show[{p1,Graphics[{Line[pts]}]}] Sum[ Sqrt[ (pts[[i 1, 1]] pts[[i, 1]])2 (pts[[i 1, 2]] pts[[i, 2]])2 ], {i, 1, n}] NIntegrate[ Sqrt[ 1 f'[x]2 ],{x, a, b}] 6.4 MOMENTS AND CENTERS OF MASS 1. Because the children are balanced, the moment of the system about the origin must be equal to zero: 5 † 80 œ x † 100 Ê x œ 4 ft, the distance of the 100-lb child from the fulcrum. 2. Suppose the log has length 2a. Align the log along the x-axis so the 100-lb end is placed at x œ a and the 200-lb end at x œ a. Then the center of mass x satisfies x œ at a distance a or
2 3
a 3
œ
2a 3
œ
" 3
(2a) which is
" 3
100(a) 200(a) 300
Ê x œ 3a . That is, x is located
of the length of the log from the 200-lb (heavier) end (see figure)
of the way from the lighter end toward the heavier end. " 3
(2a)
èëëéëëê 100 lbs. ñïïïïïïïïïïïïïïñïïïïñïïïïïïñ a 200 lbs a x œ a/3 ! 3. The center of mass of each rod is in its center (see Example 1). The rod system is equivalent to two point m masses located at the centers of the rods at coordinates ˆ L# ß !‰ and ˆ0ß L# ‰. Therefore x œ y m
œ
x" m" x# m# m" m#
œ
L # †m0
mm
œ
L 4
and y œ
mx m
œ
y" m# y# m# m" m#
œ
0 L2 †m mm
œ
L 4
Ê
ˆ L4 ß L4 ‰
is the center of
mass location. 4. Let the rods have lengths x œ L and y œ 2L. The center of mass of each rod is in its center (see Example 1). The rod system is equivalent to two point masses located at the centers of the rods at coordinates ˆ L# ß !‰ and (!ß L). Therefore x œ
L # †m0†2m
m2m
œ
5. M! œ '0 x † 4 dx œ ’4 x# “ œ 4 † 2
#
# !
6. M! œ '1 x † 4 dx œ ’4 x# “ œ 3
#
$ "
4 #
!†mL†2m m2m
L 6
and y œ
œ
4 #
œ 8; M œ '0 4 dx œ [4x]#! œ 4 † 2 œ 8 Ê x œ 2
2L 3
‰ Ê ˆ L6 ß 2L 3 is the center of mass location. M! M
œ1
(9 1) œ 16; M œ '1 4 dx œ [4x]$" œ 12 4 œ 8 Ê x œ 3
M! M
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
œ
16 8
œ2
Section 6.4 Moments and Centers of Mass 7. M! œ '0 x ˆ1 x3 ‰ dx œ '0 Šx 3
œ3
3
œ
9 6
9 #
Ê xœ
ˆ 15 ‰ 2 ˆ 92 ‰
œ
M! M
x# 3‹
œ
8. M! œ '0 x ˆ2 x4 ‰ dx œ '0 Š2x 4
4
M œ '0 ˆ2 x4 ‰ dx œ ’2x 4
9. M! œ '1 x Š1 4
" Èx ‹
% x# 8 “!
15 9
$ x$ 9 “!
#
dx œ ’ x# œ
x# 4‹
œ
27 ‰ 9
M œ '0 ˆ1 3x ‰ dx œ ’x 3
15 # ;
16 8
% x$ 12 “ !
œ ˆ16
64 ‰ 12
œ
œ
œ6 Ê xœ
dx œ '1 ˆx x"Î# ‰ dx œ ’ x# 4
#
M! M
% 2x$Î# 3 “"
32 3 †6
œ ˆ8
œ 16
16 ‰ 3
M! M
10. M! œ '1Î4 x † 3 ˆx$Î# x&Î# ‰ dx œ 3'1Î4 ˆx"Î# x$Î# ‰ dx œ 3 2x"Î# 1
1
œ 6 14 œ 20 Ê x œ
M! M
œ
2 ‘" 3x$Î# "Î%
œ
9 3
1
2
œ 3; M œ '0 (2 x) dx '1 x dx œ ’2x 1
2
" x# # “!
#
# # ’ x# “ "
œ ˆ2
12. M! œ '0 x(x 1) dx '1 2x dx œ '0 ax# xb dx '1 2x dx œ ’ x3 œ3
œ
32 3 ;
ˆ 73 ‰ 6 5
œ
2 ‘" x"Î# "Î%
œ
15 #
14 3
2
1
1
2
2
œ
23 6 ;
Ê xœ
M! M
‰ ˆ 72 ‰ œ œ ˆ 23 6
œ
73 6
;
73 30
œ 3 ’(2 2) Š2 † 16 ‰‘ 3
$
"‰ #
" x$ 3 “!
" #
2 ˆ "# ‰ ‹“
œ 3 ˆ2
14 ‰ 3
" x# 2 “!
ˆ 4#
#
$
’ x3 “ œ ˆ1 "3 ‰ ˆ 83 "3 ‰ "
"‰ #
œ3 Ê xœ
#
"
M! M
œ1
# cx# d " œ ˆ "3 2" ‰ (4 1)
M œ '0 (x 1) dx '1 2 dx œ ’ x# x“ c2xd #" œ ˆ "# 1‰ (4 #) œ 2
5 6
4528 6
œ
9 20
2
1
2 3
œ 3 ˆ2 32 ‰ ˆ4
11. M! œ '0 x(2 x) dx '1 x † x dx œ '0 a2x x# b dx '1 x# dx œ ’ 2x# 1
œ 16 †
ˆ "# 32 ‰ œ
4
œ 3(4 1) œ 9; M œ 3'1Î4 ˆx$Î# x&Î# ‰ dx œ 3 x"Î#2
16 3
16 9
% M œ '1 ˆ1 x"Î# ‰ dx œ x 2x"Î# ‘ " œ (4 4) (1 2) œ 5 Ê x œ 1
$ x# 6 “!
5 3
dx œ ’x#
œ8
œ ˆ 92
385
!
3 #
œ
7 #
23 21
13. Since the plate is symmetric about the y-axis and its density is constant, the distribution of mass is symmetric about the y-axis and the center of mass lies on the y-axis. This means that x œ 0. It remains to find y œ MMx . We model the distribution of mass with @/3-+6 strips. The typical strip has center of mass: # (µ x ßµ y ) œ Šxß x 4 ‹ , length: 4 x# , width: dx, area: #
#
dA œ a4 x b dx, mass: dm œ $ dA œ $ a4 x# b dx. The moment of the strip about the x-axis is # µ C dm œ Š x #4 ‹ $ a4 x# b dx œ #$ a16 x% b dx. The moment of the plate about the x-axis is Mx œ ' µ C dm œ 'c2 #$ a16 x% b dx œ 2
$ #
’16x
# x& 5 “ #
plate is M œ ' $ a4 x# b dx œ $ ’4x
œ
$ #
’Š16 † 2
# x$ 3 “ #
2& 5‹
œ 2$ ˆ8 83 ‰ œ
32$ 3 .
2& 5 ‹“
œ
$ †2 #
ˆ32
Therefore y œ
Mx M
œ
Š16 † 2
32 ‰ 5
$ Š 128 5 ‹
Š 323$ ‹
‰ mass is the point (xß y) œ ˆ!ß 12 5 .
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
œ
œ
128$ 5 .
12 5 .
The mass of the
The plate's center of
386
Chapter 6 Applications of Definite Integrals
14. Applying the symmetry argument analogous to the one in Exercise 13, we find x œ 0. To find y œ MMx , we use the @/3-+6 strips technique. The typical strip has center of # mass: (µ x ßµ y ) œ Šxß 25 x ‹ , length: 25 x# , width: dx, #
#
area: dA œ a25 x bdx, mass: dm œ $ dA œ $ a25 x# b dx. The moment of the strip about the x-axis is # µ y dm œ Š 25 # x ‹ $ a25 x# b dx œ
œ 'c5 #$ a25 x# b dx œ 5
#
œ $ † 625 ˆ5 œ 2$ Š5$
10 3
5$ 3‹
$ #
'c55
a25 x# b dx. The moment of the plate about the x-axis is Mx œ ' µ y dm #
$ #
$ #
a625 50x# x% b dx œ
’625x
50 3
x$
& x& 5 “ &
œ 2 † #$ Š625 † 5
50 3
† 5$
1‰ œ $ † 625 † ˆ 38 ‰ . The mass of the plate is M œ ' dm œ 'c5 $ a25 x# b dx œ $ ’25x 5
œ
$ † 5$ . Therefore y œ
4 3
Mx M
œ
$ †5% † ˆ 83 ‰ $ †5$ †ˆ 43 ‰
5& 5‹ & x$ 3 “ &
œ 10. The plate's center of mass is the point (xß y) œ (!ß 10).
15. Intersection points: x x# œ x Ê 2x x# œ 0 Ê x(2 x) œ 0 Ê x œ 0 or x œ 2. The typical @/3-+6 # strip has center of mass: (µ x ßµ y ) œ Šxß ax x b (x) ‹ #
œ Šxß
x# # ‹,
#
length: ax x b (x) œ 2x x# , width: dx,
area: dA œ a2x x# b dx, mass: dm œ $ dA œ $ a2x x# b dx. The moment of the strip about the x-axis is # µ y dm œ Š x# ‹ $ a2x x# b dx; about the y-axis it is µ x dm œ x † $ a2x x# b dx. Thus, Mx œ ' µ y dm œ '0 ˆ #$ x# ‰ a2x x# b dx œ #$ '0 a2x$ x% b dx œ #$ ’ x# 2
2
%
# x& 5 “!
œ #$ Š2$
œ 45$ ; My œ ' µ x dm œ '0 x † $ a2x x# b dx œ $ '0 a2x# x$ b œ $ ’ 23 x$ 2
2
M œ ' dm œ '0 $ a2x x# b dx œ $ '0 a2x x# b dx œ $ ’x# 2
2
œ ˆ 43$ ‰ ˆ 43$ ‰ œ 1 and y œ
Mx M
# x$ 3 “!
# x% 4 “!
2& 5‹
œ #$ † 2$ ˆ1 45 ‰
œ $ Š2 †
œ $ ˆ4 83 ‰ œ
4$ 3
2$ 3
2% 4‹
œ
. Therefore, x œ
$ †2% 1#
œ
My M
œ ˆ 45$ ‰ ˆ 43$ ‰ œ 35 Ê (xß y) œ ˆ1ß 35 ‰ is the center of mass.
16. Intersection points: x# 3 œ 2x# Ê 3x# 3 œ 0 Ê 3(x 1)(x 1) œ 0 Ê x œ 1 or x œ 1. Applying the symmetry argument analogous to the one in Exercise 13, we find x œ 0. The typical @/3-+6 strip has center of mass: # # # (µ x ßµ y ) œ Šxß 2x ax 3b ‹ œ Šxß x 3 ‹ , #
#
#
#
#
length: 2x ax 3b œ 3 a1 x b, width: dx, area: dA œ 3 a1 x# b dx, mass: dm œ $ dA œ 3$ a1 x# b dx. The moment of the strip about the x-axis is µ y dm œ 3 $ ax# 3b a1 x# b dx œ 3 $ ax% 3x# x# 3b dx œ #
œ
3 #
$ 'c1 ax% 2x# 3b dx œ 1
#
3 #
&
$ ’ x5
2x$ 3
M œ ' dm œ 3$ 'c1 a1 x# b dx œ 3$ ’x 1
3x“ " x$ 3 “ "
" "
œ
3 #
3 #
$ ax% 2x# 3b dx; Mx œ ' µ y dm
† $ † 2 ˆ 5"
2 3
45 ‰ 3‰ œ 3$ ˆ 310 œ 325$ ; 15
œ 3$ † 2 ˆ1 3" ‰ œ 4$ . Therefore, y œ
Mx M
œ 5$††$32†4 œ 58
Ê (xß y) œ ˆ0ß 85 ‰ is the center of mass.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
4$ 3
;
Section 6.4 Moments and Centers of Mass
387
17. The typical 29+6 strip has center of mass: $ (µ x ßµ y ) œ Š y y ß y‹ , length: y y$ , width: dy, #
area: dA œ ay y$ b dy, mass: dm œ $ dA œ $ ay y$ b dy. The moment of the strip about the y-axis is $ # µ x dm œ $ Š y y ‹ ay y$ b dy œ $ ay y$ b dy #
œ
$ #
#
#
%
'
ay 2y y b dy; the moment about the x-axis is
1 $ µ y dm œ $ y ay y$ b dy œ $ ay# y% b dy. Thus, Mx œ ' µ y dm œ $ '0 ay# y% b dy œ $ ’ y3
My œ ' µ x dm œ
$ #
'01 ay# 2y% y' b dy œ #$ ’ y3
$
œ $ '0 ay y$ b dy œ $ ’ y# 1
œ
#
" y% 4 “!
2y& 5
œ $ ˆ "# 4" ‰ œ
$ 4
" y( 7 “!
œ
$ #
ˆ "3
. Therefore, x œ
2 5
7" ‰ œ
$ #
œ $ ˆ "3 "5 ‰ œ
15 ‰ ˆ 35 3†42 œ 5†7
4$ ‰ ˆ 4 ‰ œ ˆ 105 $ œ
My M
" y& 5 “!
16 105
2$ 15
;
4$ 105
; M œ ' dm
Mx M
2$ ‰ ˆ 4 ‰ œ ˆ 15 $
and y œ
16 8 ‰ Ê (xß y) œ ˆ 105 ß 15 is the center of mass.
8 15
18. Intersection points: y œ y# y Ê y# 2y œ 0 Ê y(y 2) œ 0 Ê y œ 0 or y œ 2. The typical 29+6 strip has center of mass: # # (µ x ßµ y ) œ Š ay yby ß y‹ œ Š y ß y‹ , #
2
#
length: y ay yb œ 2y y# , width: dy, area: dA œ a2y y# b dy, mass: dm œ $ dA œ $ a2y y# b dy. The moment about the y-axis is µ x dm œ #$ † y# a2y y# b dy œ #$ a2y$ y% b dy; the moment about the x-axis is µ y dm œ $ y a2y y# b dy œ $ a2y# y$ b dy. Thus, Mx œ ' µ y dm œ '0 $ a2y# y$ b dy œ $ ’ 2y3 2
œ '0
2
$ #
$
a2y$ y% b dy œ
œ $ ’y#
# y$ 3 “!
$ #
%
’ y2
œ $ ˆ4 83 ‰ œ
# y& 5 “!
4$ 3
œ
$ #
ˆ8
# y% 4 “!
16$ 1#
ˆ 405 32 ‰ œ
4$ 5
; M œ ' dm œ '0 $ a2y y# b dy
œ
$ #
My M
œ ˆ 45$ ‰ ˆ 43$ ‰ œ
32 ‰ 5
. Therefore, x œ
œ
(4 3) œ
4$ 3
; My œ ' µ x dm
16 ‰ 4
œ $ ˆ 16 3
2
3 5
and y œ
Mx M
œ ˆ 43$ ‰ ˆ 43$ ‰ œ 1
Ê (xß y) œ ˆ 35 ß "‰ is the center of mass. 19. Applying the symmetry argument analogous to the one used in Exercise 13, we find x œ 0. The typical @/3-+6 strip has center of mass: (µ x ßµ y ) œ ˆxß cos# x ‰ , length: cos x, width: dx,
area: dA œ cos x dx, mass: dm œ $ dA œ $ cos x dx. The moment of the strip about the x-axis is µ y dm œ $ † cos# x † cos x dx 2x ‰ œ #$ cos# x dx œ #$ ˆ 1 cos dx œ 4$ (1 cos 2x) dx; thus, # 1Î2
Mx œ ' µ y dm œ 'c1Î2 4$ (1 cos 2x) dx œ 1Î#
œ $ [sin x]1Î# œ 2$ . Therefore, y œ
Mx M
œ
$ 4
x $1 4 †# $
œ
sin 2x ‘ 1Î# # 1Î# 1 8
œ
$ 4
ˆ 1# 0‰ ˆ 1# ‰‘ œ
$1 4
Ê (xß y) œ ˆ!ß 18 ‰ is the center of mass.
20. Applying the symmetry argument analogous to the one used in Exercise 13, we find x œ 0. The typical vertical strip has # center of mass: (µ x ßµ y ) œ Šxß sec# x ‹ , length: sec# x, width: dx, area: dA œ sec# x dx, mass: dm œ $ dA œ $ sec# x dx. The # moment about the x-axis is µ y dm œ Š sec x ‹ a$ sec# xb dx œ
$ #
1Î4
sec% x dx. Mx œ 'c1Î4 µ y dm œ
#
$ #
1Î2
; M œ ' dm œ $ '1Î2 cos x dx
'11ÎÎ44 sec% x dx
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
388
Chapter 6 Applications of Definite Integrals œ
$ #
'c11ÎÎ44 atan# x 1b asec# xb dx œ #$ '11ÎÎ44 (tan x)# asec# xb dx #$ '11ÎÎ44 sec# x dx œ 2$ ’ (tan3x) “ 1Î4
œ
$ 2
3" ˆ 3" ‰‘ #$ [1 (1)] œ
$
1Î%
1 Î4
Therefore, y œ
Mx M
œ ˆ 43$ ‰ ˆ 2"$ ‰ œ
2 3
$ 3
$ œ
4$ 3
#$ [tan x]1Î%
; M œ ' dm œ $ 'c1Î4 sec# x dx œ $ [tan x]1Î4 œ $ [1 (1)] œ 2$ . 1Î4
1Î4
Ê (xß y) œ ˆ!ß 32 ‰ is the center of mass.
21. Since the plate is symmetric about the line x œ 1 and its density is constant, the distribution of mass is symmetric about this line and the center of mass lies on it. This means that x œ 1. The typical @/3-+6 strip has center of mass: # # # (µ x ßµ y ) œ Šxß a2x x ba2x 4xb ‹ œ Šxß x 2x ‹ , #
#
#
#
#
length: a2x x b a2x 4xb œ 3x 6x œ 3 a2x x# b , width: dx, area: dA œ 3 a2x x# b dx, mass: dm œ $ dA œ 3$ a2x x# b dx. The moment about the x-axis is # µ y dm œ 3 $ ax# 2xb a2x x# b dx œ 3 $ ax# 2xb dx #
#
œ 3# $ ax% 4x$ 4x# b dx. Thus, Mx œ ' µ y dm œ '0
2
œ $ 3 2
& Š 25
%
2
4 3
$
%
†2 ‹œ $†2 3 #
œ '0 3$ a2x x# b dx œ 3$ ’x# 2
# x$ 3 “!
ˆ 25
1
2‰ 3
3 2
&
$ ax% 4x$ 4x# b dx œ 32 $ ’ x5 x% 34 x$ “ %
œ $ †2 3 #
10 ‰ ˆ 6 15 15
œ 3$ ˆ4 83 ‰ œ 4$ . Therefore, y œ
Mx M
œ
8$ 5
; M œ ' dm
# !
œ ˆ 85$ ‰ ˆ 4"$ ‰ œ 25
Ê (xß y) œ ˆ1ß 25 ‰ is the center of mass. 22. (a) Since the plate is symmetric about the line x œ y and its density is constant, the distribution of mass is symmetric about this line. This means that x œ y. The typical @/3-+6 strip has center of mass: È # (µ x ßµ y ) œ Šxß 9 x ‹ , length: È9 x# , width: dx, #
area: dA œ È9 x# dx, mass: dm œ $ dA œ $ È9 x# dx. The moment about the x-axis is È # µ y dm œ $ Š 9# x ‹ È9 x# dx œ
$ #
a9 x# b dx. Thus, Mx œ ' µ y dm œ '0
3
$ #
a9 x# b dx œ
$ #
’9x
$ x$ 3 “!
(27 9) œ 9$ ; M œ ' dm œ ' $ dA œ $ ' dA œ $ (Area of a quarter of a circle of radius 3) œ $ ˆ 941 ‰ œ 4 ‰ Therefore, y œ MMx œ (9$ ) ˆ 91$ œ 14 Ê (xß y) œ ˆ 14 ß 14 ‰ is the center of mass. œ
$ #
(b) Applying the symmetry argument analogous to the one used in Exercise 13, we find that x œ 0. The typical vertical strip has the same parameters as in part (a). 3 Thus, M œ ' µ y dm œ ' $ a9 x# b dx x
œ #'0
3
$ #
c3 #
a9 x# b dx œ 2(9$ ) œ 18$ ;
M œ ' dm œ ' $ dA œ $ ' dA œ $ (Area of a semi-circle of radius 3) œ $ ˆ 921 ‰ œ 91$ 2 . Therefore, y œ 4 as in part (a) Ê (xß y) œ ˆ0ß 1 ‰ is the center of mass.
Mx M
2 ‰ œ (18$ ) ˆ 91$ œ
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
4 1
, the same y
91$ 4
.
Section 6.4 Moments and Centers of Mass
389
23. Since the plate is symmetric about the line x œ y and its density is constant, the distribution of mass is symmetric about this line. This means that x œ y. The typical @/3-+6 strip has È # center of mass: (µ x ßµ y ) œ Šxß 3 9 x ‹ , #
length: 3 È9 x# , width: dx, area: dA œ Š3 È9 x# ‹ dx, mass: dm œ $ dA œ $ Š3 È9 x# ‹ dx. The moment about the x-axis is µ y dm œ $
Š3 È9 x# ‹ Š3 È9 x# ‹ #
dx œ
$ #
c9 a9 x# bd dx œ
$ x# #
dx. Thus, Mx œ '0
3
$ x# #
dx œ
$ 6
$
cx$ d ! œ #
9$ #
equals the area of a square with side length 3 minus one quarter the area of a disk with radius 3 Ê A œ 3 œ
9 4
9$ 4
(4 1) Ê M œ $ A œ
(4 1). Therefore, y œ
Mx M
œ ˆ 9#$ ‰ ’ 9$(44 1) “ œ
2 41
. The area 19 4
Ê (xß y) œ ˆ 4 2 1 ß 4 2 1 ‰ is the
center of mass. 24. Applying the symmetry argument analogous to the one used in Exercise 13, we find that y œ 0. The typical @/3-+6 strip has center of mass: (µ x ßµ y ) œ Œxß " x$
length:
ˆ x"$ ‰ œ
2 x$
" x$
x"$ #
œ (xß 0),
, width: dx, area: dA œ
2 x$
dx,
2$ x$
mass: dm œ $ dA œ dx. The moment about the y-axis is a µ x dm œ x † 2x$$ dx œ 2x$# dx. Thus, My œ ' µ x dm œ '1 2x$# dx œ 2$ x" ‘ " œ 2$ ˆ "a 1‰ œ a
xœ
My M
œ
’ 2$(aa 1) “
25. Mx œ ' µ y dm œ '1
2
# ’ $ aa#a 1b “
Š x2# ‹ #
2$ (a1) a
œ
2a a1
; M œ ' dm œ '1
a
Ê (xß y) œ
2$ x$
ˆ a 2a ‰ 1ß 0 .
dx œ $ x"# ‘ " œ $ ˆ a"# 1‰ œ a
$ aa# 1b a#
. Therefore,
Also, a lim x œ 2. Ä_
† $ † ˆ x2# ‰ dx
œ '1 ˆ x"# ‰ ax# b ˆ x2# ‰ dx œ '1 2
2
2 x#
dx œ 2'1 x# dx 2
#
œ 2 cx" d " œ 2 ˆ "# ‰ (1)‘ œ 2 ˆ "# ‰ œ 1;
My œ ' µ x dm œ '1 x † $ † ˆ x2# ‰ dx 2
œ '1 x ax# b ˆ x2# ‰ dx œ 2'1 x dx œ 2 ’ x# “ 2
2
#
# "
œ 2 ˆ2 "# ‰ œ 4 1 œ 3; M œ ' dm œ '1 $ ˆ x2# ‰ dx œ '1 x# ˆ x2# ‰ dx œ 2'1 dx œ 2[x]"# œ 2(2 1) œ 2. So xœ
My M
œ
3 #
and y œ
Mx M
œ
" #
2
2
2
Ê (xß y) œ ˆ 3# ß "# ‰ is the center of mass.
26. We use the @/3-+6 strip approach: 1 # M œ'µ y dm œ ' ax x b ax x# b † $ dx x
œ
0
" #
#
'0 ax# x% b † 12x dx 1
œ 6'0 ax$ x& b dx œ 6 ’ x4 1
œ 6 ˆ "4 6" ‰ œ
%
6 4
1œ
" #
" x' 6 “!
;
My œ ' µ x dm œ '0 x ax x# b † $ dx œ '0 ax# x$ b † 12x dx œ 12'0 ax$ x% b dx œ 12 ’ x4 1
1
1
%
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
" x& 5 “!
œ 12 ˆ "4 5" ‰
390
Chapter 6 Applications of Definite Integrals œ
12 #0
xœ
œ
; M œ ' dm œ ' ax x# b † $ dx œ 12'0 ax# x$ b dx œ 12 ’ x3 1
3 5
1
$
0
My M
œ
3 5
and y œ
Mx M
œ
" #
" x% 4 “!
œ 12 ˆ "3 4" ‰ œ
12 12
œ 1. So
Ê ˆ 35 ß "# ‰ is the center of mass.
shell ‰ shell 27. (a) We use the shell method: V œ 'a 21 ˆ radius Š height ‹ dx œ '1 21x ’ È4x Š È4x ‹“ dx b
œ 161'1
4
x Èx
4
% dx œ 161'1 x"Î# dx œ 161 32 x$Î# ‘ " œ 161 ˆ 32 † 8 32 ‰ œ 4
(b) Since the plate is symmetric about the x-axis and its density $ (x) œ
" x
321 3
(8 1) œ
2241 3
is a function of x alone, the
distribution of its mass is symmetric about the x-axis. This means that y œ 0. We use the vertical strip 4 4 4 approach to find x: My œ ' µ x dm œ '1 x † ’ È4x Š È4x ‹“ † $ dx œ '1 x † È8x † x" dx œ 8'1 x"Î# dx 4 œ 8 2x"Î# ‘ " œ 8(2 † 2 2) œ 16; M œ ' dm œ '1 ’ È4x Š È ‹“ † $ dx œ 8'1 Š È"x ‹ ˆ "x ‰ dx œ 8'1 x$Î# dx x %
4
%
œ 8 2x"Î# ‘ " œ 8[1 (2)] œ 8. So x œ
My M
4
œ
4
œ 2 Ê (xß y) œ (2ß 0) is the center of mass.
16 8
(c)
28. (a) We use the disk method: V œ 'a 1R# (x) dx œ '1 1 ˆ x4# ‰ dx œ 41'1 x# dx œ 41 x" ‘ " b
4
4
%
‘ œ 41 " 4 (1) œ 1[1 4] œ 31
(b) We model the distribution of mass with vertical strips: Mx œ ' µ y dm œ '1
4
2 œ 2'1 x$Î# dx œ 2 ’ È x dm œ '1 x † “ œ 2[1 (2)] œ 2; My œ ' µ x %
4
$Î#
4
"
%
2‘ œ 2 ’ 2x3 “ œ 2 16 3 3 œ "
œ 2(4 2) œ 4. So x œ
My M
28 3
œ
; M œ ' dm œ '1
4
ˆ 28 ‰ 3 4
œ
7 3
and y œ
2 x
Mx M
† $ dx œ 2'1
4
œ
2 4
œ
" #
Èx x
2 x
ˆ 2x ‰ 2
† ˆ 2x ‰ † $ dx œ '1
4
2 x#
† Èx dx
† $ dx œ 2'1 x"Î# dx 4
dx œ 2'1 x"Î# dx œ 2 2x"Î# ‘ " %
4
Ê (xß y) œ ˆ 73 ß "# ‰ is the center of mass.
(c)
29. The mass of a horizontal strip is dm œ $ dA œ $ L dy, where L is the width of the triangle at a distance of y above its base on the x-axis as shown in the figure in the text. Also, by similar triangles we have Ê Lœ
b h
(h y). Thus, Mx œ ' µ y dm œ '0 $ y ˆ bh ‰ (h y) dy œ h
œ
$b h
Š h#
$
h$ 3‹
œ
$b h
#
h# 2‹
Šh
œ $ bh# ˆ "# 3" ‰ œ œ
$ bh 2
. So y œ
Mx M
$ bh# 6
œ
$b h
# ˆ 2 ‰ Š $bh 6 ‹ $ bh
œ
h 3
œ
hy h
'0h ahy y# b dy œ $hb ’ hy#
; M œ ' dm œ '0 $ ˆ hb ‰ (h y) dy œ h
L b
$b h
#
h
y$ 3 “!
'0h ah yb dy œ $hb ’hy y2 “ h
Ê the center of mass lies above the base of the
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
#
!
Section 6.4 Moments and Centers of Mass triangle one-third of the way toward the opposite vertex. Similarly the other two sides of the triangle can be placed on the x-axis and the same results will occur. Therefore the centroid does lie at the intersection of the medians, as claimed. 30. From the symmetry about the y-axis it follows that x œ 0. It also follows that the line through the points (!ß !) and (!ß $) is a median Ê y œ "3 (3 0) œ 1 Ê (xß y) œ (!ß ").
31. From the symmetry about the line x œ y it follows that x œ y. It also follows that the line through the points (!ß !) and ˆ "# ß "# ‰ is a median Ê y œ x œ 23 † ˆ "# 0‰ œ 3" Ê (xß y) œ ˆ "3 ß 3" ‰ . 32. From the symmetry about the line x œ y it follows that x œ y. It also follows that the line through the point (!ß !) and ˆ #a ß #a ‰ is a median Ê y œ x œ 32 ˆ #a 0‰ œ 3" a Ê (xß y) œ ˆ 3a ß 3a ‰ . 33. The point of intersection of the median from the vertex (0ß b) to the opposite side has coordinates ˆ!ß #a ‰ Ê y œ (b 0) † "3 œ b3 and x œ ˆ #a !‰ † 32 œ 3a Ê (xß y) œ ˆ 3a ß b3 ‰ .
34. From the symmetry about the line x œ
a #
it follows that
xœ It also follows that the line through the points a ˆ # ß !‰ and ˆ #a ß b‰ is a median Ê y œ "3 (b 0) œ b3 a #.
Ê (xß y) œ ˆ #a ß b3 ‰ .
35. y œ x"Î# Ê dy œ
" #
x"Î# dx
Ê ds œ È(dx)# (dy)# œ É1 Mx œ $ '0 Èx É1 2
œ $ '0 Éx 2
dx œ
" ‰$Î# 4
œ
2$ 3
œ
2$ ˆ 9 ‰$Î# 3 ’ 4
’ˆ2
" 4
ˆ 4" ‰
" 4x
" 4x
dx
2$ 3
$Î# ’ˆx 4" ‰ “
dx ;
# !
ˆ 4" ‰$Î# “ $Î#
“œ
2$ 3
"‰ ˆ 27 8 8 œ
13$ 6
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
391
392
Chapter 6 Applications of Definite Integrals
36. y œ x$ Ê dy œ 3x# dx Ê dx œ É(dx)# a3x# dxb# œ È1 9x% dx; Mx œ $ '0 x$ È1 9x% dx; 1
" 36
[u œ 1 9x% Ê du œ 36x$ dx Ê
du œ x$ dx;
x œ 0 Ê u œ 1, x œ 1 Ê u œ 10] Ä Mx œ $ '1
10
" 36
u"Î# du œ
$ 36
23 u$Î# ‘ "! œ "
$ 54
ˆ10$Î# 1‰
37. From Example 6 we have Mx œ '0 a(a sin ))(k sin )) d) œ a# k'0 sin# ) d) œ 1
œ
a# k #
)
sin 2) ‘ 1 # !
œ
a# k 1 #
1
'01 (1 cos 2)) d)
; My œ '0 a(a cos ))(k sin )) d) œ a# k '0 sin ) cos ) d) œ 1
1
M œ '0 ak sin ) d) œ ak[ cos )]1! œ 2ak. Therefore, x œ 1
a# k #
My M
œ 0 and y œ
Mx M
a# k #
1
csin# )d ! œ 0;
#
" ‰ œ Š a 2k1 ‹ ˆ 2ak œ
a1 4
Ê ˆ!ß a41 ‰
is the center of mass. 38. Mx œ ' µ y dm œ '0 (a sin )) † $ † a d) 1
œ '0 aa# sin )b a1 k kcos )kb d) 1
œ a# '0 (sin ))(1 k cos )) d) 1Î2
a# '1Î2 (sin ))(1 k cos )) d) 1
œ a# '0 sin ) d) a# k'0 sin ) cos ) d) a# '1Î2 sin ) d) a# k '1Î2 sin ) cos ) d) 1Î2
1Î2
1Î#
#
1
a# k ’ sin# ) “
œ a# [ cos )]!
1Î# !
1
#
a# [ cos )]11Î# a# k ’ sin# ) “
1 1Î#
œ a [0 (1)] a k ˆ "# 0‰ a# [(1) 0] a# k ˆ0 "# ‰ œ a# #
#
a# k #
a#
œ 2a# a# k œ a# (2 k); 1 1 M œ'µ x dm œ ' (a cos )) † $ † a d) œ ' aa# cos )b a1 k kcos )kb d) y
0
0
œa
'0
œ a#
'01Î2 cos ) d) a# k '
#
1Î2
#
(cos ))(1 k cos )) d) a 1Î2
0
#
œ a [sin
1Î# ) ]!
œ a# (1 0)
a# k #
a# k #
a# k #
)
'1Î2 (cos ))(1 k cos )) d) 1
2) ‰ 2) ‰ ˆ 1 cos d) a# '1Î2 cos ) d) a# k'1Î2 ˆ 1 cos d) # #
sin 2) ‘ 1Î# # !
1
a# [sin )]11Î#
1
a# k #
ˆ 1# 0‰ (! 0)‘ a# (0 1)
)
a# k #
sin 2) ‘ 1 # 1Î#
(1 0) ˆ 1# 0‰‘ œ a#
a# k 1 4
a#
M œ '0 $ † a d) œ a'0 (1 k kcos )k) d) œ a '0 (1 k cos )) d) a'1Î2 (1 k cos )) d) 1
1
1Î#
œ a[) k sin )]! œ
a1 #
1Î2
a# k 1 4
œ 0;
1
a[) k sin )]11Î# œ a ˆ 1# k‰ 0‘ a (1 0) ˆ 1# k‰‘
ak a ˆ 1# k‰ œ a1 2ak œ a(1 2k). So x œ
My M
œ 0 and y œ
Mx M
œ
a# (2 k) a(1 #k)
œ
a(2 k) 1 #k
ka ‰ Ê ˆ0ß 2a 1 #k is the center of mass.
39. Consider the curve as an infinite number of line segments joined together. From the derivation of arc length we have that the length of a particular segment is ds œ È(dx)# (dy)# . This implies that Mx œ ' $ y ds, My œ ' $ x ds and M œ ' $ ds. If $ is constant, then x œ yœ
Mx M
' y ds
œ ' ds œ
' y ds length
My M
' x ds
œ ' ds œ
' x ds length
and
.
40. Applying the symmetry argument analogous to the one used in Exercise 13, we find that x œ 0. The typical
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 6.5 Areas of Surfaces of Revolution and the Theorems of Pappus x#
a vertical strip has center of mass: (µ x ßµ y ) œ Œxß 2 4p , length: a
mass: dm œ $ dA œ $ Ša œ
$ #
2
%
&
#
œ $ ’ax 8a$ Èpa 3
2
Èpa œ 2 † $ ’ax È Mx M
œŠ
Èpa
2
Èpa
x$ 12p “ !
œ
3 5
Èpa
8a# $Èpa 5
2$ paÈpa 12p ‹
œ 2$ Š2aÈpa
8a# $ Èpa 3 ‹ Š 8a$È 5 pa ‹
2
x& 80p# “ 0
œ 2 † #$ ’a# x
"6 ‰ ‰ œ 2a# $ Èpa ˆ 8080 œ 2a# $ Èpa ˆ 64 80 œ
x$ 12p “ c2 pa
. So y œ
c2Èpa
#
16 ‰ 80
width: dx, area: dA œ Ša
dx. Thus, Mx œ ' µ y dm œ 'c2Èpa "# Ša
#Èpa x x 'c22ÈÈpapa Ša# 16p ‹ dx œ #$ ’a# x 80p “
œ 2a# $ Èpa ˆ1
œ
x# 4p ‹
x# 4p ,
x# 4p ‹ Ša
x# 4p ‹ $
œ 4a$ Èpa ˆ1
dx,
dx 2& p# a# Èpa ‹ 80p#
œ $ Š2a# Èpa
; M œ ' dm œ $
x# 4p ‹
2
Èpa
'
c2Èpa
4 ‰ 12
Ša
x# 4p ‹
dx
œ 4a$ Èpa ˆ 121#4 ‰
a, as claimed.
41. Since the density is constant, its value will not affect our answers, so we can set $ œ ". 1Î2 ! A generalization of Example 6 yields M œ ' µ y dm œ ' a# sin ) d) œ a# [ cos )]1Î2 ! 1Î2 !
1Î# !
x
1Î2 !
œ a# cos ˆ 1# !‰ cos ˆ 1# !‰‘ œ a# (sin ! sin !) œ 2a# sin !; M œ ' dm œ '1Î# ! a d) œ a[)]11ÎÎ22 !! œ a ˆ 1# !‰ ˆ 1# !‰‘ œ 2a!. Thus, y œ Ê c œ 2a sin !. Then y œ
a(2a sin !) 2a!
œ
ac s ,
Mx M
œ
2a# sin ! 2a!
œ
a sin ! !
lim
! Ä !b
(b)
sin ! ! cos ! ! ! cos !
! f(!)
¸
d h
œ
a sin ! ! sin ! ! cos ! ! ! cos ! .
a(sin ! ! cos !) a(! ! cos !)
œ
œ a cos ! d Ê d œ
0.4 0.664879
0.6 0.662615
0.8 0.659389
1.0 0.655145
6.5 AREAS OF SURFACES OF REVOLUTION AND THE THEOREMS OF PAPPUS 1. (a)
dy dx
#
% œ sec# x Ê Š dy dx ‹ œ sec x
Ê S œ 21'0
1Î4
a(sin ! ! cos !) . !
The graphs below suggest that
2 3.
0.2 0.666222
c #
as claimed.
42. (a) First, we note that y œ (distance from origin to AB) d Ê Moreover, h œ a a cos ! Ê
. Now s œ a(2!) and a sin ! œ
(b)
(tan x) È1 sec% x dx
(c) S ¸ 3.84
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
393
394 2. (a)
Chapter 6 Applications of Definite Integrals dy dx
#
(b)
2 œ 2x Ê Š dy dx ‹ œ 4x
Ê S œ 21'0 x# È1 4x# dx 2
(c) S ¸ 53.23
3. (a) xy œ 1 Ê x œ Ê S œ 21'1
2
" y
" y
Ê
dx dy
#
œ y"# Ê Š dx dy ‹ œ
" y%
(b)
È1 y% dy
(c) S ¸ 5.02
4. (a)
dx dy
#
# œ cos y Ê Š dx dy ‹ œ cos y
(b)
Ê S œ 21'0 (sin y) È1 cos# y dy 1
(c) S ¸ 14.42
# 5. (a) x"Î# y"Î# œ 3 Ê y œ ˆ3 x"Î# ‰ "Î# ‰ ˆ ˆ Ê dy "# x"Î# ‰ dx œ 2 3 x #
"Î# ‰ ˆ Ê Š dy dx ‹ œ 1 3x
(b)
#
# # Ê S œ 21'1 ˆ3 x"Î# ‰ É1 a1 3x"Î# b dx 4
(c) S ¸ 63.37
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 6.5 Areas of Surfaces of Revolution and the Theorems of Pappus dx dy
6. (a)
#
"Î# ‰ ˆ œ 1 y"Î# Ê Š dx dy ‹ œ 1 y
#
395
(b)
# Ê S œ 21 '1 ˆy 2Èy‰ É1 a1 y"Î# b dx 2
(c) S ¸ 51.33
dx dy
7. (a)
#
(b)
# œ tan y Ê Š dx dy ‹ œ tan y
Ê S œ 21'0 Š'0 tan t dt‹ È1 tan# y dy 1Î3
y
œ 21'0 Š'0 tan t dt‹ sec y dy 1Î3
y
(c) S ¸ 2.08
dy dx
8. (a)
#
(b)
# œ Èx# 1 Ê Š dy dx ‹ œ x 1
È5
Ê S œ 21'1 Š'1 Èt# 1 dt‹ È1 ax# 1b dx
È5
x
œ 21'1 Š'1 Èt# 1 dt‹ x dx x
(c) S ¸ 8.55
9. y œ œ
x #
1È5 #
Ê
dy dx
' ˆ x ‰ É1 œ "# ; S œ 'a 21y Ê1 Š dy dx ‹ dx Ê S œ 0 21 #
x #
4
" 4
dx œ
1È5 #
'04 x dx
%
# ’ x# “ œ 41È5; Geometry formula: base circumference œ 21(2), slant height œ È4# 2# œ 2È5
!
Ê Lateral surface area œ
10. y œ
#
b
Ê x œ 2y Ê
dx dy
" #
(41) Š2È5‹ œ 41È5 in agreement with the integral value
# È È ' È # ' œ 2; S œ 'c 21x Ê1 Š dx dy ‹ dy œ 0 21 † 2y 1 2 dy œ 41 5 0 y dy œ 21 5 cy d ! #
d
2
2
#
œ 21È5 † 4 œ 81È5; Geometry formula: base circumference œ 21(4), slant height œ È4# 2# œ 2È5 Ê Lateral surface area œ " (81) Š2È5‹ œ 81È5 in agreement with the integral value #
11.
dy dx
' œ "# ; S œ 'a 21yÊ1 Š dy dx ‹ dx œ 1 21 b
#
3
(x 1) #
É1 ˆ "# ‰# dx œ
1È5 #
'13 (x 1) dx
œ
1È5 #
1È5 #
#
’ x# x“
$ "
È ˆ 9# 3‰ ˆ "# 1‰‘ œ 1 # 5 (4 2) œ 31È5; Geometry formula: r" œ "# "# œ 1, r# œ 3# "# œ 2, œ slant height œ È(2 1)# (3 1)# œ È5 Ê Frustum surface area œ 1(r" r# ) ‚ slant height œ 1(1 2)È5
œ 31È5 in agreement with the integral value
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
396
Chapter 6 Applications of Definite Integrals
12. y œ
x #
" #
Ê x œ 2y 1 Ê
dx œ 2; S œ 'c 21x Ê1 Š dy ‹ dy œ '1 21(2y 1)È1 4 dy œ 21È5 '1 (2y 1) dy #
d
dx dy
2
#
2
œ 21È5 cy# yd " œ 21È5 [(4 2) (1 1)] œ 41È5; Geometry formula: r" œ 1, r# œ 3, slant height œ È(2 1)# (3 1)# œ È5 Ê Frustum surface area œ 1(1 3)È5 œ 41È5 in agreement with the integral value 13.
dy dx
#
x# 3
œ
’u œ 1
x% 9
Ê S œ '0
2
x% 9
Ê Š dy dx ‹ œ Ê du œ
4 9
x œ 0 Ê u œ 1, x œ 2 Ê u œ Ä S œ 21 '1
25Î9
14.
œ
1 3
dy dx
œ
" 1 2 4 du œ # 3 1 ˆ 12527 ‰ œ 98811 3 27 #
Ê S œ '3Î4 21Èx É1 œ 21'3Î4 Éx 15Î4
15.
œ
’ˆ 15 4
œ
41 3
(8 1) œ
dy dx
" (2 2x) # È2x x#
œ
ˆ 43
dx;
dx;
#&Î*
u$Î# ‘ "
" 4x
dx
$Î# dx œ 21 ’ 23 ˆx 4" ‰ “
" 4
" ‰$Î# 4
41 3
x$ 9
du œ
x% 9
" 4x
x"Î# Ê Š dy dx ‹ œ 15Î4
É1
25 ‘ 9
u"Î# †
ˆ 125 ‰ 27 1 œ " #
" 4
x$ dx Ê
21 x$ 9
" ‰$Î# “ 4
"&Î% $Î%
41 3
$ ’ˆ 24 ‰
Ê Š dy dx ‹ œ
(1 x)# 2x x#
œ
1“
281 3
œ
#
1x È2x x#
Ê S œ '0 5 21È2x x# É1 1Þ5 Þ
œ 21'0 5 È2x 1Þ5 Þ
(1 x)# 2x x#
È x# 1 2x x# x# 2x È 2x x#
dx
dx
œ 21'0 5 dx œ 21[x]"Þ& !Þ& œ 21 1Þ5 Þ
16.
dy dx
" 2È x 1
œ
#
dy Ê Š dx ‹ œ
" 4(x 1)
Ê S œ '1 21Èx 1 É1 5
œ 21'1 É(x 1) 5
" 4
" 4(x 1)
dx
dx œ 21'1 Éx 5
&
$Î# œ 21 ’ 23 ˆx 54 ‰ “ œ
17.
œ
41 3
œ
1 6
dx dy
‰$Î# ’ˆ 25 4
5 4
41 ˆ 5 ‰$Î# 3 ’ 5 4 " $ $ ˆ 94 ‰$Î# “ œ 431 Š 52$ 32$ ‹
(125 27) œ
981 6
œ
dx ˆ1 45 ‰$Î# “
491 3
% ' œ y# Ê Š dx dy ‹ œ y Ê S œ 0 #
1
u œ 1 y% Ê du œ 4y$ dy Ê
" 4
21 y$ 3
È1 y% dy;
du œ y$ dy; y œ 0
Ê u œ 1, y œ 1 Ê u œ 2d Ä S œ '1 21 ˆ "3 ‰ u"Î# ˆ 4" du‰ 2
œ
1 6
'12 u"Î# du œ 16 32 u$Î# ‘ #" œ 19 ŠÈ8 1‹
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 6.5 Areas of Surfaces of Revolution and the Theorems of Pappus 18. x œ ˆ "3 y$Î# y"Î# ‰ Ÿ 0, when 1 Ÿ y Ÿ 3. To get positive area, we take x œ ˆ "3 y$Î# y"Î# ‰ Ê
dx dy
#
œ "# ˆy"Î# y"Î# ‰ Ê Š dx dy ‹ œ
" 4
ay 2 y" b
Ê S œ '1 21 ˆ "3 y$Î# y"Î# ‰ É1 4" ay 2 y" b dy 3
œ 21'1 ˆ 3" y$Î# y"Î# ‰ É 4" ay 2 y" b dy 3
Éay"Î# y"Î# b#
œ 21'1 ˆ "3 y$Î# y"Î# ‰ 3
œ 1'1 ˆ "3 y# 3
2 3
dx dy
œ
" È 4 y
œ 41 '0
15Î4
œ
20.
dx dy
81 3
œ
3
$
y 1‰ dy œ 1 ’ y9
œ 19 (18 1 3) œ 19.
dy œ 1'1 y"Î# ˆ 3" y 1‰ Šy"Î#
#
#
Ê Š dx dy ‹ œ
Ê S œ '0
15Î4
" 4 y
5È 5 8 ‹
œ
81 3
#
" È2y 1
Ê Š dx dy ‹ œ
Š 40
È 5 5 È 5 ‹ 8
œ
È
È5
1 12
‹œ
#
41 È 2 3
" 2y 1
$Î#
’1$Î# ˆ 58 ‰
“œ
#
1 dy œ ÊŠy'
" #
" 16y' ‹
2
È2
È2
dy dx
œ
" #
aa# x# b
Ê S œ 21'ca Èa# x# É1 a
$Î#
œ
21 r h
dy dx
#
#
É h h# r
25. y œ cos x Ê
œ
41 È 2 3
5È 5 ‹ 8È 8
Š1
21 40
" #
" 16y' ‹
dy
dy œ 21'1 ˆy% "4 y# ‰ dy 2
" 4y$ ‹
(8 † 31 5) œ
2531 20
È2
x ax# 1b dx œ 21'0 ax$ xb dx œ 21 ’ x4
"Î#
x# aa # x # b
(2x) œ
x È a# x#
%
#
Ê Š dy dx ‹ œ
È# x# # “!
œ 21 ˆ 44 22 ‰ œ 41
x# aa # x # b
dx œ 21'ca Èaa# x# b x# dx œ 21'ca a dx œ 21a[x]ca a a
a
r h
#
Ê Š dy dx ‹ œ
r# h#
Ê S œ 21 '0
h
r h
x É1
r# h#
dx œ 21'0
h
r h
#
#
x É h h# r dx
'0h x dx œ 2h1r Èh# r# ’ x# “ h œ 2h1r Èh# r# Š h# ‹ œ 1rÈh# r# #
#
dy dx
5$Î# “
1
œ 21a[a (a)] œ (21a)(2a) œ 41a# x Ê
1‰
È2
23. y œ Èa# x# Ê
r h
" 3
Ê dy œ xÈx# 2 dx Ê ds œ È1 a2x# x% b dx Ê S œ 21'0 x È1 2x# x% dx
$Î#
œ 21'0 xÉax# 1b# dx œ 21'0
24. y œ
dy œ 21'5Î8 È(2y 1) 1 dy
2
#
ax# 2b
" 9
È(4 y) 1 dy
5$Î# “ œ 831 ’ˆ 45 ‰
dy; S œ '1 21y ds œ 21'1 y Šy$
" 4y$ ‹
"
" 3
1‰‘ œ 1 ˆ3
1 dy œ ÊŠy'
"‰ "‰ ˆ " " ‰‘ œ 21 ˆ 31 œ 21 ’ y5 4" y" “ œ 21 ˆ 32 5 8 5 4 5 8 œ
22. y œ
15Î4
15 ‰$Î# 4
1
" 4y$ ‹
dy œ Šy$
&
dy œ 41'0
" 3
Š16È2 5È5‹
21. ds œ Èdx# dy# œ ÊŠy$ " 4y$ ‹
3
351È5 3
"
5 Š 8†2 8†22È 2
dy œ 1 '1 ˆ 3" y 1‰ (y 1) dy
3‰ ˆ "9
9 3
" 4y
21 † 2È4 y É1
Ê S œ '5Î8 21È2y 1 É1
" 2y1
1
œ ÊŠy$
"
È5 y dy œ 41 23 (5 y)$Î# ‘ "&Î% œ 831 ’ˆ5 !
Š 5È 5
41 È 2 3
y“ œ 1 ˆ 27 9
161 9
œ 21'5Î8 È2 y"Î# dy œ 21È2 23 y$Î# ‘ &Î) œ œ
$
y# 3
" ‹ y"Î#
#
0
#
# È1 sin# x dx ' œ sin x Ê Š dy dx ‹ œ sin x Ê S œ 21 c1Î2 (cos x) #
1Î2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
397
398
Chapter 6 Applications of Definite Integrals
26. y œ ˆ1 x#Î$ ‰
$Î#
Ê
dy dx
œ
Ê S œ 2'0 21 ˆ1 x#Î$ ‰ 1
3 #
ˆ1 x#Î$ ‰"Î# ˆ 23 x"Î$ ‰ œ
$Î#
#
ˆ1x#Î$ ‰"Î# x"Î$
Ê Š dy dx ‹ œ
1x#Î$ x#Î$
œ
" x#Î$
1
$Î# " É1 ˆ x#Î$ 1‰ dx œ 41'0 ˆ1 x#Î$ ‰ Èx#Î$ dx 1
$Î# œ 41'0 ˆ1 x#Î$ ‰ x"Î$ dx; u œ 1 x#Î$ Ê du œ 23 x"Î$ dx Ê 32 du œ x"Î$ dx; 1
! x œ 0 Ê u œ 1, x œ 1 Ê u œ 0d Ä S œ 41'1 u$Î# ˆ 3# du‰ œ 61 25 u&Î# ‘ " œ 61 ˆ0 25 ‰ œ 0
121 5
# # # È16# y# 27. The area of the surface of one wok is S œ 'c 21x Ê1 Š dx dy ‹ dy. Now, x y œ 16 Ê x œ #
d
Ê
dx dy
œ c7
#
y È16# y#
Ê Š dx dy ‹ œ
c7
; S œ 'c16 21È16# y# É1
y# 16# y#
y# 16# y#
c7
dy œ 21'c16 Èa16# y# b y# dy
œ 21'c16 16 dy œ 321 † 9 œ 2881 ¸ 904.78 cm# . The enamel needed to cover one surface of one wok is
V œ S † 0.5 mm œ S † 0.05 cm œ (904.78)(0.05) cm$ œ 45.24 cm$ . For 5000 woks, we need 5000 † V œ 5000 † 45.24 cm$ œ (5)(45.24)L œ 226.2L Ê 226.2 liters of each color are needed. 28. y œ Èr# x# Ê œ 21'a
abh
œ 21'a
2x È r# x #
œ
Èar# x# b x# dx œ 21r' a
29. y œ ÈR# x# Ê abh
œ "#
dy dx
dy dx
œ "#
2x È R # x#
x Èr# x#
abh
œ
abh
30. (a) x# y# œ 45# Ê x œ È45# y# Ê 45
x# r# x # ;
S œ 21 'a
abh
Èr# x# É1
x# r# x#
dx
dx œ 21rh, which is independent of a. #
x È R # x#
ÈaR# x# b x# dx œ 21R ' a
S œ 'c22Þ5 21 È45# y# É1
#
Ê Š dx dy ‹ œ
y# 45# y#
dx Ê Š dy ‹ œ
x# R # x# ;
S œ 21'a
abh
ÈR# x# É1
x# R # x#
dx
dx œ 21Rh dx dy
œ
y È45# y#
#
Ê Š dx dy ‹ œ
y# 45# y#
;
dy œ 21 '22Þ5 Èa45# y# b y# dy œ 21 † 45'22Þ5 dy 45
45
œ (21)(45)(67.5) œ 60751 square feet (b) 19,085 square feet dy ' È1 1 dx œ 21 ' (x)È2 dx 21' xÈ2 dx 31. y œ x Ê Š dy dx ‹ œ 1 Ê Š dx ‹ œ 1 Ê S œ 21 c1 kxk c1 0 #
# œ 2È21 ’ x# “
32.
dy dx
œ
x# 3
!
2
#
"
!
#
4 9
0
# 2È21 ’ x# “ œ 2È21 ˆ0 "# ‰ 2È21(2 0) œ 5È21
Ê Š dy dx ‹ œ
Ê du œ
2
x% 9
Ê by symmetry of the graph that S œ 2 'cÈ3 21 Š x9 ‹ É1 0
$
x% 9
dx; ’u œ 1
x$ dx Ê "4 du œ x9 dx; x œ È3 Ê u œ 2, x œ 0 Ê u œ 1“ Ä S œ 41'2 u"Î# ˆ "4 ‰ du 1
$
" œ 1'2 u"Î# du œ 1 23 u$Î# ‘ # œ 1 Š 23 23 È8‹ œ 1
È3
È3
21 3
ŠÈ8 1‹ . If the absolute value bars are dropped the
integral for S œ 'cÈ3 21f(x) ds will equal zero since 'cÈ3 21 Š x9 ‹ É1 $
x% 9
dx is the integral of an odd function
over the symmetric interval È3 Ÿ x Ÿ È3.
33.
dx dt
x% 9
œ sin t and
dy dt
#
È( sin t)# (cos t)# œ 1 Ê S œ ' 21y ds ‰ Š dy œ cos t Ê Êˆ dx dt dt ‹ œ #
œ '0 21(2 sin t)(1) dt œ 21 c2t cos td #!1 œ 21[(41 1) (0 1)] œ 81# 21
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 6.5 Areas of Surfaces of Revolution and the Theorems of Pappus 34.
dx dt
œ t"Î# and
È3
dy dt
œ '0 21 ˆ 23 t$Î# ‰ É t
#
" t
dt œ
È3
'0
#
21 ˆ 23 t$Î# ‰ É t #
f(t) œ 21 ˆ 23 t$Î# ‰ É t Ê
35.
dx dt
È3
'0
F(t) dt œ
œ 1 and
È2
dy dt
#
281 9
" t
41 3
È3
'0
1 t
Ê S œ ' 21x ds
tÈt# 1 dt; cu œ t# 1 Ê du œ 2t dt; t œ 0 Ê u œ 1,
'14 231 Èu du œ 491 u$Î# ‘ %" œ 2891
’t œ È3 Ê u œ 4“ Ä Note:
#
#
Èt t" œ É t ‰ Š dy œ t"Î# Ê Êˆ dx dt dt ‹ œ
1 t
dt is an improper integral but limb f(t) exists and is equal to 0, where tÄ!
. Thus the discontinuity is removable: define F(t) œ f(t) for t 0 and F(0) œ 0
. #
#
# È2‹ œ Ét# 2È2 t 3 Ê S œ ' 21x ds ‰ Š dy œ t È2 Ê Êˆ dx dt dt ‹ œ Ê1 Št #
œ 'cÈ2 21 Št È2‹ Ét# 2È2 t 3 dt; ’u œ t# 2È2 t 3 Ê du œ Š2t 2È2‹ dt; t œ È2 Ê u œ 1,
* t œ È2 Ê u œ 9“ Ä '1 1Èu du œ 23 1u$Î# ‘ " œ 9
36.
dx dt
œ aa1 cos tb and
dy dt
21 3
(27 1) œ
521 3
#
#
‰ Š dy Éc aa1 cos tb d# aa sin tb# œ a sin t Ê Êˆ dx dt dt ‹ œ
œ Èa2 2 a2 cos t a2 cos2 t a2 sin2 t œ È2a2 2a2 cos t œ aÈ2È1 cos t Ê S œ ' 21y ds œ '0 21 aa1 cos tb † aÈ2È1 cos t dt œ 2È2 1 a2 '0 a1 cos tb3/2 dt 21
37.
dx dt
œ 2 and
21
dy dt
È2# 1# œ È5 Ê S œ ' 21y ds œ ' 21(t 1)È5 dt ‰ Š dy œ 1 Ê Êˆ dx dt dt ‹ œ 0 #
#
1
"
#
œ 21È5 ’ t2 t“ œ 31È5. Check: slant height is È5 Ê Area is 1(1 2)È5 œ 31È5 . !
38.
dx dt
œ h and
dy dt
Èh# r# Ê S œ ' 21y ds œ ' 21rtÈh# r# dt ‰ Š dy œ r Ê Êˆ dx dt dt ‹ œ 0
œ 21rÈh# r#
#
#
1
'01 t dt œ 21rÈh# r# ’ t2 “ " œ 1rÈh# r# . #
!
Check: slant height is Èh# r# Ê Area is
1rÈh# r# . 39. (a) An equation of the tangent line segment is (see figure) y œ f(mk ) f w (mk )(x mk ). When x œ xkc1 we have r" œ f(mk ) f w (mk )(x51 mk ) œ f(mk ) f w (mk ) ˆ ?#xk ‰ œ f(mk ) f w (mk ) when x œ xk we have r# œ f(mk ) f w (mk )(x5 mk ) k œ f(mk ) f w (mk ) ?x # ;
(b) L#k œ (?xk )# (r# r" )#
;
#
ˆf w (mk ) ?#xk ‰‘ œ (?xk )# [f w (mk )?xk ]# Ê Lk œ È(?xk )# [f w (mk )?xk ]# , as claimed (c) From geometry it is a fact that the lateral surface area of the frustum obtained by revolving the tangent œ (?xk )# f w (mk )
?x k #
?x k #
line segment about the x-axis is given by ?Sk œ 1(r" r# )Lk œ 1[2f(mk )] Éa?xk b# [f w (mk )?xk ]# using parts (a) and (b) above. Thus, ?Sk œ 21f(mk ) È1 [f w (mk )]# ?xk .
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399
400
Chapter 6 Applications of Definite Integrals
! ?Sk œ lim ! 21f(mk ) È1 [f w (mk )]# ?xk œ ' 21f(x) È1 [f w (x)]# dx (d) S œ n lim Ä_ nÄ_ a kœ1 kœ1 n
n
È3
40. S œ 'a 21f(x) dx œ '0 b
21 †
x È3
b
dx œ
1 È3
È3
c x# d ! œ
31 È3
œ È31
41. The centroid of the square is located at (#ß #). The volume is V œ (21) ayb (A) œ (21)(2)(8) œ 321 and the surface area is S œ (21) ayb (L) œ (21)(2) Š4È8‹ œ 32È21 (where È8 is the length of a side). 42. The midpoint of the hypotenuse of the triangle is ˆ 3# ß 3‰ Ê y œ 2x is an equation of the median Ê the line y œ 2x contains the centroid. The point ˆ 3# ß $‰ is 3È 5 #
units from the origin Ê the x-coordinate of the
# centroid solves the equation Ɉx 3# ‰ (2x 3)#
œ
È5 #
Ê ˆx# 3x 94 ‰ a4x# 12x 9b œ
5 4
Ê 5x# 15x 9 œ 1 Ê x# 3x 2 œ (x 2)(x 1) œ 0 Ê x œ 1 since the centroid must lie inside the triangle Ê y œ 2. By the Theorem of Pappus, the volume is V œ (distance traveled by the centroid)(area of the region) œ 21 a5 xb "# (3)(6)‘ œ (21)(4)(9) œ 721
43. The centroid is located at (#ß !) Ê V œ (21) axb (A) œ (21)(2)(1) œ 41# 44. We create the cone by revolving the triangle with vertices (0ß 0), (hß r) and (hß 0) about the x-axis (see the accompanying figure). Thus, the cone has height h and base radius r. By Theorem of Pappus, the lateral surface area swept out by the hypotenuse L is given by S œ 21yL œ 21 ˆ r ‰ Èh# r# #
œ 1rÈr# h# . To calculate the volume we need the position of the centroid of the triangle. From the diagram we see that the centroid lies on the line y œ œ
" 3
Éh#
r# 4
#
r 2h
#
# x. The x-coordinate of the centroid solves the equation É(x h)# ˆ 2hr x #r ‰ #
#
Ê Š 4h4h# r ‹ x# Š 4h 2h r ‹ x
inside the triangle Ê y œ
r 2h
47. V œ 21 yA Ê
4 3
2 ar# 4h# b 9
œ0 Ê xœ
2h 3
or
4h 3
Ê xœ
x œ 3r . By the Theorem of Pappus, V œ 21 ˆ 3r ‰‘ ˆ "# hr‰ œ
45. S œ 21 y L Ê 41a# œ a21yb (1a) Ê y œ 46. S œ 213 L Ê 21 ˆa
r# 4
2a ‰‘ (1a) 1
2a 1,
since the centroid must lie
1r# h.
and by symmetry x œ 0
œ 21a# (1 2)
1ab# œ a21yb ˆ 1#ab ‰ Ê y œ
48. V œ 213A Ê V œ 21 ˆa
" 3
2h 3 ,
4a ‰‘ 1a# Š # ‹ 31
œ
4b 31
and by symmetry x œ 0
1a$ (31 4) 3
49. V œ 213 A œ (21)(area of the region) † (distance from the centroid to the line y œ x a). We must find the 4a ‰ distance from ˆ0ß 31 to y œ x a. The line containing the centroid and perpendicular to y œ x a has slope 1 and contains the point ˆ!ß 34a1 ‰ . This line is y œ x 34a1 . The intersection of y œ x a and y œ x 34a1 is the point ˆ 4a 613a1 ß 4a 613a1 ‰ . Thus, the distance from the centroid to the line y œ x a is Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 6.6 Work Ɉ 4a 613a1 ‰# ˆ 34a1
4a 61
3a1 ‰# 61
œ
È2 (4a 3a1) 61
Ê V œ (21) Š
È2 (4a 3a1) # ‹ Š 1#a ‹ 61
œ
È2 1a$ (4 31) 6
‰ 50. The line perpendicular to y œ x a and passing through the centroid ˆ!ß 2a 1 has equation y œ x intersection of the two perpendicular lines occurs when x a œ x
Ê xœ
2a 1
2a a1 21
Ê yœ
2a ‰# #
a(21) È 21
#
the distance from the centroid to the line y œ x a is Ɉ 2a 2 1a 0‰ ˆ 2a 2 1a
œ
2a 1 . The 2a a1 21 . Thus
.
1 ) Therefore, by the Theorem of Pappus the surface area is S œ 21 ’ a(2 “ (1a) œ È21a# (2 1). È 21
51. From Example 4 and Pappus's Theorem for Volumes we have the moment about the x-axis is Mx œ y M #
œ ˆ 34a1 ‰ Š 1#a ‹ œ
2a$ 3
.
6.6 WORK 1. The force required to stretch the spring from its natural length of 2 m to a length of 5 m is F(x) œ kx. The work done by F is W œ '0 F(x) dx œ k '0 x dx œ 3
3
k #
$
cx# d ! œ
9k # .
This work is equal to 1800 J Ê
k œ 1800
9 #
Ê k œ 400 N/m 2. (a) We find the force constant from Hooke's Law: F œ kx Ê k œ
Ê kœ
F x
800 4
œ 200 lb/in.
(b) The work done to stretch the spring 2 inches beyond its natural length is W œ '0 kx dx 2
œ 200 '0 x dx œ 200 ’ x# “ œ 200(2 0) œ 400 in † lb œ 33.3 ft † lb 2
#
# !
(c) We substitute F œ 1600 into the equation F œ 200x to find 1600 œ 200x Ê x œ 8 in. 3. We find the force constant from Hooke's law: F œ kx. A force of 2 N stretches the spring to 0.02 m N Ê 2 œ k † (0.02) Ê k œ 100 m . The force of 4 N will stretch the rubber band y m, where F œ ky Ê y œ Ê yœ
4N N 100 m
œ 100 '0
0Þ04
Ê y œ 0.04 m œ 4 cm. The work done to stretch the rubber band 0.04 m is W œ '0
F k
0Þ04
#
x dx œ 100 ’ x# “
!Þ!%
œ
!
(100)(0.04)# #
kx dx
œ 0.08 J
4. We find the force constant from Hooke's law: F œ kx Ê k œ
F x
Ê kœ
90 1
Ê k œ 90
N m. &
The work done to
‰ stretch the spring 5 m beyond its natural length is W œ '0 kx dx œ 90 '0 x dx œ 90 ’ x# “ œ (90) ˆ 25 # œ 1125 J 5
5
#
!
5. (a) We find the spring's constant from Hooke's law: F œ kx Ê k œ
F x
œ
21,714 8 5
œ
21,714 3
Ê k œ 7238
(b) The work done to compress the assembly the first half inch is W œ '0 kx dx œ 7238 '0 0Þ5
#
œ 7238 ’ x# “
!Þ& !
#
œ (7238) (0.5) # œ
(7238)(0.25) #
1Þ0
1Þ0
Þ
Þ
#
¸ 2714 in † lb 6. First, we find the force constant from Hooke's law: F œ kx Ê k œ compresses the scale x œ scale this far is W œ '0
1Î8
in, he/she must weigh F œ kx œ #
kx dx œ 2400 ’ x# “
"Î) !
x dx
¸ 905 in † lb. The work done to compress the assembly the
second half inch is: W œ '0 5 kx dx œ 7238 '0 5 x dx œ 7238 ’ x# “
" 8
lb in
0Þ5
œ
2400 2†64
F x
2,400 ˆ 8" ‰
"Þ! !Þ&
œ
œ
150 " ‰ ˆ 16
7238 #
c1 (0.5)# d œ
(7238)(0.75) #
œ 16 † 150 œ 2,400
lb in .
If someone
œ 300 lb. The work done to compress the
œ 18.75 lb † in. œ
25 16
ft † lb
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401
402
Chapter 6 Applications of Definite Integrals
7. The force required to haul up the rope is equal to the rope's weight, which varies steadily and is proportional to x, the length of the rope still hanging: F(x) œ 0.624x. The work done is: W œ '0 F(x) dx œ '0 0.624x dx 50
#
œ 0.624 ’ x# “
&! !
50
œ 780 J
8. The weight of sand decreases steadily by 72 lb over the 18 ft, at 4 lb/ft. So the weight of sand when the bag is x ft off the ground is Faxb œ "%% %x. The work done is: W œ 'a F(x) dx œ '0 a"%% %xbdx œ c144x 2x# d ! œ 1944 ft † lb b
18
")
9. The force required to lift the cable is equal to the weight of the cable paid out: F(x) œ (4.5)(180 x) where x is the position of the car off the first floor. The work done is: W œ '0
180
œ 4.5 ’180x
")! x# # “!
180# # ‹
œ 4.5 Š180#
œ
4.5†180# #
F(x) dx œ 4.5'0
180
(180 x) dx
œ 72,900 ft † lb
10. Since the force is acting toward the origin, it acts opposite to the positive x-direction. Thus F(x) œ xk# . The b work done is W œ 'a xk# dx œ k 'a x"# dx œ k x" ‘ a œ k ˆ b" "a ‰ œ b
b
k(a b) ab
11. The force against the piston is F œ pA. If V œ Ax, where x is the height of the cylinder, then dV œ A dx Ê Work œ ' F dx œ ' pA dx œ 'ap
ap# ßV# b " ßV" b
p dV.
12. pV"Þ% œ c, a constant Ê p œ cV"Þ% . If V" œ 243 in$ and p" œ 50 lb/in$ , then c œ (50)(243)"Þ% œ 109,350 lb. ‘ ˆ 3#"!Þ% Thus W œ '243 109,350V"Þ% dV œ 109,350 œ 109,350 0.4 0.4V!Þ% #%$ $#
32
" ‰ #43!Þ%
ˆ 4" 9" ‰ œ 109,350 0.4
œ (109,350)(5) (0.4)(36) œ 37,968.75 in † lb. Note that when a system is compressed, the work done by the system is negative. 13. Let r œ the constant rate of leakage. Since the bucket is leaking at a constant rate and the bucket is rising at a constant rate, the amount of water in the bucket is proportional to a#! xb, the distance the bucket is being raised. The leakage rate of the water is 0.8 lb/ft raised and the weight of the water in the bucket is F œ 0.8a#! xb. So: W œ '0 0.8a#! xb dx œ 0.8 ’20x 20
#! x# # “!
œ 160 ft † lb.
14. Let r œ the constant rate of leakage. Since the bucket is leaking at a constant rate and the bucket is rising at a constant rate, the amount of water in the bucket is proportional to a#! xb, the distance the bucket is being raised. The leakage rate of the water is 2 lb/ft raised and the weight of the water in the bucket is F œ 2a#! xb. So: W œ '0 2a#! xb dx œ 2 ’20x 20
#! x# # “!
œ 400 ft † lb.
Note that since the force in Exercise 14 is 2.5 times the force in Exercise 13 at each elevation, the total work is also 2.5 times as great.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 6.6 Work
403
15. We will use the coordinate system given. (a) The typical slab between the planes at y and y ?y has a volume of ?V œ (10)(12) ?y œ 120 ?y ft$ . The force F required to lift the slab is equal to its weight: F œ 62.4 ?V œ 62.4 † 120 ?y lb. The distance through which F must act is about y ft, so the work done lifting the slab is about ?W œ force ‚ distance œ 62.4 † 120 † y † ?y ft † lb. The work it takes to lift all 20
the water is approximately W ¸ ! ?W 0 20
œ ! 62.4 † 120y † ?y ft † lb. This is a Riemann sum for 0
the function 62.4 † 120y over the interval 0 Ÿ y Ÿ 20. The work of pumping the tank empty is the limit of these sums: W œ '0 62.4 † 120y dy œ (62.4)(120) ’ y# “ 20
#
#! !
‰ œ (62.4)(120)(200) œ 1,497,600 ft † lb œ (62.4)(120) ˆ 400 #
5 ‰ (b) The time t it takes to empty the full tank with ˆ 11 –hp motor is t œ
W †lb 250 ftsec
œ
1,497,600 ft†lb †lb 250 ftsec
œ 5990.4 sec
œ 1.664 hr Ê t ¸ 1 hr and 40 min (c) Following all the steps of part (a), we find that the work it takes to lower the water level 10 ft is W œ '0 62.4 † 120y dy œ (62.4)(120) ’ y# “ 10
#
œ 1497.6 sec œ 0.416 hr ¸ 25 min (d) In a location where water weighs 62.26
"! !
‰ œ 374,400 ft † lb and the time is t œ œ (62.4)(120) ˆ 100 #
W †lb 250 ftsec
lb ft$ :
a) W œ (62.26)(24,000) œ 1,494,240 ft † lb. b) t œ 1,494,240 œ 5976.96 sec ¸ 1.660 hr Ê t ¸ 1 hr and 40 min 250 In a location where water weighs 62.59
lb ft$
a) W œ (62.59)(24,000) œ 1,502,160 ft † lb b) t œ 1,502,160 œ 6008.64 sec ¸ 1.669 hr Ê t ¸ 1 hr and 40.1 min 250 16. We will use the coordinate system given. (a) The typical slab between the planes at y and y ?y has a volume of ?V œ (20)(12) ?y œ 240 ?y ft$ . The force F required to lift the slab is equal to its weight: F œ 62.4 ?V œ 62.4 † 240 ?y lb. The distance through which F must act is about y ft, so the work done lifting the slab is about ?W œ force ‚ distance 20
œ 62.4 † 240 † y † ?y ft † lb. The work it takes to lift all the water is approximately W ¸ ! ?W 10 20
œ ! 62.4 † 240y † ?y ft † lb. This is a Riemann sum for the function 62.4 † 240y over the interval 10
10 Ÿ y Ÿ 20. The work it takes to empty the cistern is the limit of these sums: W œ '10 62.4 † 240y dy 20
#
œ (62.4)(240) ’ y# “ (b) t œ
W †lb 275 ftsec
œ
#!
œ (62.4)(240)(200 50) œ (62.4)(240)(150) œ 2,246,400 ft † lb
"! 2,246,400 ft†lb 275
¸ 8168.73 sec ¸ 2.27 hours ¸ 2 hr and 16.1 min
(c) Following all the steps of part (a), we find that the work it takes to empty the tank halfway is W œ '10 62.4 † 240y dy œ (62.4)(240) ’ y# “ 15
#
Then the time is t œ
W †lb 275 ftsec
œ
936,000 #75
"& "!
œ (62.4)(240) ˆ 225 #
100 ‰ #
‰ œ 936,000 ft. œ (62.4)(240) ˆ 125 #
¸ 3403.64 sec ¸ 56.7 min
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
404
Chapter 6 Applications of Definite Integrals lb ft$ :
(d) In a location where water weighs 62.26
a) W œ (62.26)(240)(150) œ 2,241,360 ft † lb. b) t œ 2,241,360 œ 8150.40 sec œ 2.264 hours ¸ 2 hr and 15.8 min 275 ‰ œ 933,900 ft † lb; t œ 933,900 c) W œ (62.26)(240) ˆ 125 # #75 œ 3396 sec ¸ 0.94 hours ¸ 56.6 min lb ft$
In a location where water weighs 62.59
a) W œ (62.59)(240)(150) œ 2,253,240 ft † lb. b) t œ 2,253,240 œ 8193.60 sec œ 2.276 hours ¸ 2 hr and 16.56 min 275 ‰ œ 938,850 ft † lb; t œ 938,850 c) W œ (62.59)(240) ˆ 125 # 275 ¸ 3414 sec ¸ 0.95 hours ¸ 56.9 min #
17. The slab is a disk of area 1x# œ 1ˆ y# ‰ , thickness ˜y, and height below the top of the tank a"! yb. So the work to pump #
the oil in this slab, ˜W, is 57a"! yb1ˆ y# ‰ . The work to pump all the oil to the top of the tank is W œ '0
10
571 # 4 a"!y
y$ bdy œ
571 4
$
’ "!$y
"! y% % “!
œ 11,8751 ft † lb ¸ 37,306 ft † lb. #
18. Each slab of oil is to be pumped to a height of 14 ft. So the work to pump a slab is a"% yba1bˆ y# ‰ and since the tank is half full and the volume of the original cone is V œ "$ 1r# h œ "$ 1a&# ba"!b œ with half the volume the cone is filled to a height y, œ
571 "%y$ 4 ’ $
$ È &!! y% “ % !
#&!1 '
#&!1 $
ft3 , half the volume œ $ È &!!
$ œ $" 1 y% y Ê y œ È &!! ft. So W œ '0 #
#&!1 '
ft3 , and
571 # 4 a"%y
y$ b dy
¸ 60,042 ft † lb. #
‰ ?y 19. The typical slab between the planes at y and and y ?y has a volume of ?V œ 1(radius)# (thickness) œ 1 ˆ 20 # œ 1 † 100 ?y ft$ . The force F required to lift the slab is equal to its weight: F œ 51.2 ?V œ 51.2 † 1001 ?y lb Ê F œ 51201 ?y lb. The distance through which F must act is about (30 y) ft. The work it takes to lift all the 30
30
kerosene is approximately W ¸ ! ?W œ ! 51201(30 y) ?y ft † lb which is a Riemann sum. The work to pump the 0
0
tank dry is the limit of these sums: W œ '0 51201(30 y) dy œ 51201 ’30y 30
¸ 7,238,229.48 ft † lb
$! y# # “!
‰ œ (5120)(4501) œ 51201 ˆ 900 #
20. (Alternate Solution) Each method must pump all of the water the 15 ft to the base of the tank. Pumping to the rim requires all the water to be pumped an additional 6 feet. Pumping into the bottom requires that the water be pumped an average of 3 additional feet. Thus pumping through the valve requires È$ fta%1b6 ft3 a'#Þ% lb/ft3 b ¸ 14,115 ft † lb less work and thus less time. 21. (a) Follow all the steps of Example 5 but make the substitution of 64.5 W œ '0
8
œ
64.51 4
64.51†8$ 3
(10 y)y# dy œ
64.51 4
$
’ 10y 3
%
y 4
)
“ œ !
64.51 4
$
Š 103†8
lb ft$
8% 4‹
for 57
lb ft$ .
Then,
1‰ ‰ œ ˆ 64.5 a8$ b ˆ 10 4 3 2
œ 21.51 † 8$ ¸ 34,582.65 ft † lb
(b) Exactly as done in Example 5 but change the distance through which F acts to distance ¸ (13 y) ft. Then W œ '0
8
571 4
(13 y)y# dy œ
571 4
œ (191) a8# b (7)(2) ¸ 53.482.5 ft † lb
$
’ 13y 3
) y% 4 “!
œ
571 4
$
Š 133†8
8% 4‹
‰ œ ˆ 5741 ‰ a8$ b ˆ 13 3 2 œ
571†8$ †7 3 †4
22. The typical slab between the planes of y and y?y has a volume of about ?V œ 1(radius)# (thickness) # œ 1 ˆÈy‰ ?y œ xy ?y m$ . The force F(y) is equal to the slab's weight: F(y) œ 10,000 mN$ † ?V œ 110,000y ?y N. The height of the tank is 4# œ 16 m. The distance through which F(y) must act to lift the slab to the level of the top of the tank is about (16 y) m, so the work done lifting the slab is about
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 6.6 Work
405
?W œ 10,0001y(16 y) ?y N † m. The work done lifting all the slabs from y œ 0 to y œ 16 to the top is 16
approximately W ¸ ! 10,0001y(16 y)?y. Taking the limit of these Riemann sums, we get 0
W œ '0 10,0001y(16 y) dy œ 10,0001'0 a16y y# b dy œ 10,0001 ’ 16y # 16
œ
16
10,000†1†16$ 6
#
"' y$ 3 “!
$
œ 10,0001 Š 16#
16$ 3 ‹
¸ 21,446,605.9 J
23. The typical slab between the planes at y and y?y has a volume of about ?V œ 1(radius)# (thickness) # œ 1 ˆÈ25 y# ‰ ?y m$ . The force F(y) required to lift this slab is equal to its weight: F(y) œ 9800 † ?V #
œ 98001 ˆÈ25 y# ‰ ?y œ 98001 a25 y# b ?y N. The distance through which F(y) must act to lift the slab to the level of 4 m above the top of the reservoir is about (4 y) m, so the work done is approximately ?W ¸ 98001 a25 y# b (4 y) ?y N † m. The work done lifting all the slabs from y œ 5 m to y œ 0 m is 0
approximately W ¸ ! 98001 a25 y# b (4 y) ?y N † m. Taking the limit of these Riemann sums, we get c5
W œ 'c5 98001 a25 y# b (4 y) dy œ 98001 'c5 a100 25y 4y# y$ b dy œ 98001 ’100C 0
0
25†25 #
œ 98001 ˆ500
† 125
4 3
625 ‰ 4
25 #
y# 34 y$
¸ 15,073,099.75 J
24. The typical slab between the planes at y and y?y has a volume of about ?V œ 1(radius)# (thickness) # œ 1 ˆÈ100 y# ‰ ?y œ 1 a100 y# b ?y ft$ . The force is F(y) œ 56ft$lb † ?V œ 561 a100 y# b ?y lb. The distance through which F(y) must act to lift the slab to the level of 2 ft above the top of the tank is about (12 y) ft, so the work done is ?W ¸ 561 a100 y# b (12 y) ?y lb † ft. The work done lifting all the slabs 10
from y œ 0 ft to y œ 10 ft is approximately W ¸ ! 561 a100 y# b (12 y) ?y lb † ft. Taking the limit of these 0
Riemann sums, we get W œ '0 561 a100 y b (12 y) dy œ 561'0 a100 y# b (12 y) dy 10
10
#
œ 561'0 a1200 100y 12y# y$ b dy œ 561 ’1200C 10
œ 561 ˆ12,000
10,000 #
4 † 1000
10,000 ‰ 4
100y# #
12y$ 3
"! y% 4 “!
œ (561) ˆ12 5 4 5# ‰ (1000) ¸ 967,611 ft † lb.
It would cost (0.5)(967,611) œ 483,805¢ œ $4838.05. Yes, you can afford to hire the firm. 25. F œ m œ
" #
dv dt
œ mv
by the chain rule Ê W œ 'x mv x#
dv dx #
"
m cv# (x# ) v (x" )d œ
26. weight œ 2 oz œ
" #
weight 32
œ
" 8
3#
œ
" #56
28. weight œ 1.6 oz œ 0.1 lb Ê m œ " 8
x# "
dv ‰ dx
dx œ m "# v# (x)‘ x" x#
" slugs; W œ ˆ "# ‰ ˆ #56 slugs‰ (160 ft/sec)# ¸ 50 ft † lb
hr 1 min 5280 ft 27. 90 mph œ 901 hrmi † 601 min † 60 sec † 1 mi œ 132 ft/sec; m œ 0.3125 lb ‰ # W œ ˆ "# ‰ ˆ 32 ft/sec# (132 ft/sec) ¸ 85.1 ft † lb
29. weight œ 2 oz œ
dx œ m'x ˆv
mv## "# mv"# , as claimed.
lb; mass œ
2 16
dv dx
lb Ê m œ
" 8
32
0.1 lb 32 ft/sec#
slugs œ
œ " #56
" 3 #0
0.3125 lb 32 ft/sec#
œ
0.3125 32
slugs;
slugs; W œ ˆ "# ‰ ˆ 3"#0 slugs‰ (280 ft/sec)# œ 122.5 ft † lb
slugs; 124 mph œ
(124)(5280) (60)(60)
¸ 181.87 ft/sec;
" W œ ˆ "# ‰ ˆ 256 slugs‰ (181.87 ft/sec)# ¸ 64.6 ft † lb
30. weight œ 14.5 oz œ 31. weight œ 6.5 oz œ
14.5 16
6.5 16
lb Ê m œ
lb Ê m œ
14.5 (16)(32)
6.5 (16)(32)
"4.5 slugs; W œ ˆ "# ‰ Š (16)(32) slugs‹ (88 ft/sec)# ¸ 109.7 ft † lb
6.5 slugs; W œ ˆ "# ‰ Š (16)(32) slugs‹ (132 ft/sec)# ¸ 110.6 ft † lb
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
! y% 4 “ &
406
Chapter 6 Applications of Definite Integrals
32. F œ (18 lb/ft)x Ê W œ '0 18x dx œ c9x# d ! 1Î6
mœ
" 8
32
œ
" #56
"Î6
slugs and v" œ 0 ft/sec. Thus,
1 4
œ
1 4
ft † lb. Now W œ È2 4
È2 4 ‹
(16) Š
È2 4 ‹
1 4
ft † lb,
sec when the bearing is at the top of its path. È2 4
The height the bearing reaches is s œ 8È2 t 16t# Ê at t œ Š8È2‹ Š
mv# "# mv"# , where W œ
" ft † lb. œ ˆ #" ‰ ˆ #56 slugs‰ v# Ê v œ 8È2 ft/sec. With v œ 0
at the top of the bearing's path and v œ 8È2 32t Ê t œ #
" #
the bearing reaches a height of
œ 2 ft
33. (a) From the diagram, rayb œ '! x œ '! É&!# ay 325b# for 325 Ÿ y Ÿ 375 ft. (b) The volume of a horizontal slice of the funnel # is ˜V ¸ 1rayb‘ ˜y #
œ 1”'! É&!# ay 325b# • ˜y (c) The work required to lift the single slice of water is ˜W ¸ 62.4˜Va$(& yb #
œ 62.4a$(& yb1”'! É&!# ay 325b# • ˜y. The total work to pump our the funnel is W #
œ '325 62.4a375 yb1”'! É50# ay 325b# • dy 375
¸ 6.3358 † 10( ft † lb. 34. (a) From the result in Example 6, the work to pump out the throat is 1,353,869,354 ft † lb. Therefor, the total work required to pump out the throat and the funnel is 1,353,869,354 63,358,000 œ 1,417227,354 ft † lb. (b) In horsepower-hours, the work required to pump out the glory hole is 1,417227,354 œ 715.8. Therefore, it would take 1.98†106 715.8 hp†h 1000 hp
œ 0.7158 hours ¸ 43 minutes.
35. We imagine the milkshake divided into thin slabs by planes perpendicular to the y-axis at the points of a partition of the interval [!ß (]. The typical slab between the planes at y and y ?y has a volume of about #
17.5 ‰ ?V œ 1(radius)# (thickness) œ 1 ˆ y14 ?y in$ . The force F(y) required to lift this slab is equal to its
weight: F(y) œ
4 9
?V œ
41 9
#
17.5 ‰ ˆ y14 ?y oz. The distance through which F(y) must act to lift this slab to
the level of 1 inch above the top is about (8 y) in. The work done lifting the slab is about 17.5)# ?W œ ˆ 491 ‰ (y14 (8 y) ?y in † oz. The work done lifting all the slabs from y œ 0 to y œ 7 is # 7
approximately W œ ! 0
41 9†14#
(y 17.5)# (8 y) ?y in † oz which is a Riemann sum. The work is the limit of
these sums as the norm of the partition goes to zero: W œ '0
7
œ
41 9†14#
œ
41 9†14#
'07 a2450 26.25y 27y# y$ b dy œ 9†4141 ’
7% 4
$
9†7
26.25 #
41 9†14#
%
#
(y 17.5)# (8 y) dy
’ y4 9y$
26.25 #
y# 2450y“
( !
#
† 7 2450 † 7“ ¸ 91.32 in † oz
36. We fill the pipe and the tank. To find the work required to fill the tank follow Example 6 with radius œ 10 ft. Then ?V œ 1 † 100 ?y ft$ . The force required will be F œ 62.4 † ?V œ 62.4 † 1001 ?y œ 62401 ?y lb. The distance through which F must act is y so the work done lifting the slab is about ?W" œ 62401 † y † ?y lb † ft. The work it takes to
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 6.7 Fluid Pressures and Forces 385
385
360
360
lift all the water into the tank is: W" ¸ ! ?W" œ ! 62401 † y † ?y lb † ft. Taking the limit we end up with W" œ '360 62401y dy œ 62401 ’ y# “ 385
#
$)& $'!
62401 #
œ
c385# 360# d ¸ 182,557,949 ft † lb 4 #
To find the work required to fill the pipe, do as above, but take the radius to be Then ?V œ 1 †
" 36
$
?y ft and F œ 62.4 † ?V œ
integration: W# ¸ ! ?W# Ê W# œ '0 360
360
0
62.4 36
62.41 36
" 6
in œ
ft.
?y. Also take different limits of summation and
1y dy œ
62.41 36
#
$'!
#
1 ‰ 360 œ ˆ 62.4 Š # ‹ ¸ 352,864 ft † lb. 36
’ y# “
!
The total work is W œ W" W# ¸ 182,557,949 352,864 ¸ 182,910,813 ft † lb. The time it takes to fill the W tank and the pipe is Time œ 1650 ¸ 182,910,813 ¸ 110,855 sec ¸ 31 hr 1650 37. Work œ '6 370 000
35ß780ß000 ß
1000 MG r#
ß
dr œ 1000 MG '6 370 000
35ß780ß000 ß
ß
" œ (1000) a5.975 † 10#% b a6.672 † 10"" b Š 6,370,000
$&ß()!ß!!!
œ 1000 MG "r ‘ 'ß$(!ß!!!
dr r#
" 35,780,000 ‹
¸ 5.144 ‚ 10"! J
38. (a) Let 3 be the x-coordinate of the second electron. Then r# œ (3 1)# Ê W œ 'c1 F(3) d3 0
œ 'c1 a23(3‚101)# b d3 œ ’ 233‚10" “
#* !
#*
0
"
œ a23 ‚ 10#* b ˆ1 #" ‰ œ 11.5 ‚ 10#*
(b) W œ W" W# where W" is the work done against the field of the first electron and W# is the work done against the field of the second electron. Let 3 be the x-coordinate of the third electron. Then r#" œ (3 1)# and r## œ (3 1)# Ê W" œ '3
5
œ a23 ‚ 10
b ˆ "4
#*
"‰ #
œ
23 4
23‚10#* r#"
‚ 10
d3 œ '3
#*
5
23‚10#* (3 ")#
, and W# œ '
d3 œ 23 ‚ 10#* ’ 3 " " “
23‚10#* r## 3
&
œ 23 ‚ 10#* ’ 3 " " “ œ a23 ‚ 10#* b ˆ 6" 4" ‰ œ $
#* ‰ #* ‰ W œ W" W# œ ˆ 23 ˆ 23 œ 4 ‚ 10 12 ‚ 10
23 3
5
23‚10#* 12
d3 œ '
23‚10#* # 3 (3 " ) 5
(3 2) œ
23 12
& $
d3
‚ 10#* . Therefore
‚ 10#* ¸ 7.67 ‚ 10#* J
6.7 FLUID PRESSURES AND FORCES 1. To find the width of the plate at a typical depth y, we first find an equation for the line of the plate's right-hand edge: y œ x 5. If we let x denote the width of the right-hand half of the triangle at depth y, then x œ 5 y and the total width is L(y) œ 2x œ 2(5 y). The depth of the strip is (y). The force exerted by the c2
c2
water against one side of the plate is therefore F œ 'c5 w(y) † L(y) dy œ 'c5 62.4 † (y) † 2(5 y) dy c2
œ 124.8 'c5 a5y y# b dy œ 124.8 5# y# "3 y$ ‘ & œ 124.8 ˆ 5# † 4 #
œ (124.8) ˆ 105 #
117 ‰ 3
" 3
† 8‰ ˆ 5# † 25
" 3
† 125‰‘
œ (124.8) ˆ 315 6 234 ‰ œ 1684.8 lb
2. An equation for the line of the plate's right-hand edge is y œ x 3 Ê x œ y 3. Thus the total width is L(y) œ 2x œ 2(y 3). The depth of the strip is (2 y). The force exerted by the water is F œ 'c3 w(2 y)L(y) dy œ 'c3 62.4 † (2 y) † 2(3 y) dy œ 124.8'c3 a6 y y# b dy œ 124.8 ’6y 0
0
œ (124.8) ˆ18
9 #
0
y# #
! y$ 3 “ $
‰ 9‰ œ (124.8) ˆ 27 # œ 1684.8 lb
3. Using the coordinate system of Exercise 4, we find the equation for the line of the plate's right-hand edge is y œ x 3 Ê x œ y 3. Thus the total width is L(y) œ 2x œ 2(y 3). The depth of the strip changes to (4 y) Ê F œ 'c3 w(4 y)L(y) dy œ 'c3 62.4 † (4 y) † 2(y 3) dy œ 124.8'c3 a12 y y# b dy 0
œ 124.8 ’12y
0
y# #
! y$ “ 3 $
œ (124.8) ˆ36
0
9 #
‰ 9‰ œ (124.8) ˆ 45 # œ 2808 lb
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
407
408
Chapter 6 Applications of Definite Integrals
4. Using the coordinate system of Exercise 4, we see that the equation for the line of the plate's right-hand edge remains the same: y œ x 3 Ê x œ 3 y and L(y) œ 2x œ 2(y 3). The depth of the strip changes to (y) Ê F œ 'c3 w(y)L(y) dy œ 'c3 62.4 † (y) † 2(y 3) dy œ 124.8'c3 ay# 3yb dy œ 124.8 ’ y3 3# y# “ 0
0
œ (124.8) ˆ 27 3
œ
27 ‰ #
0
(124.8)(27)(2 3) 6
$
! $
œ 561.6 lb
5. Using the coordinate system of Exercise 4, we find the equation for the line of the plate's right-hand edge to be y œ 2x 4 Ê x œ y # 4 and L(y) œ 2x œ y 4. The depth of the strip is (1 y). (a) F œ 'c4 w(1 y)L(y) dy œ 'c4 62.4 † (1 y)(y 4) dy œ 62.4 'c4 a4 3y y# b dy œ 62.4 ’4y 0
0
œ (62.4) ’(4)(4)
(3)(16) #
(b) F œ (64.0) ’(4)(4)
0
(3)(16) #
64 3 “
œ (62.4) ˆ16 24
64 3 “
œ
(64.0)(120 64) 3
64 ‰ 3
œ
(62.4)(120 64) 3
3y# #
! y$ 3 “ %
œ 1164.8 lb
¸ 1194.7 lb
6. Using the coordinate system given, we find an equation for the line of the plate's right-hand edge to be y œ 2x 4 Ê x œ 4#y and L(y) œ 2x œ 4 y. The depth of the strip is (1 y) Ê F œ '0 w(1 y)(4 y) dy 1
œ 62.4'0 ay# 5y 4b dy œ 62.4 ’ y3 1
$
œ (62.4) ˆ "3
5 #
5y# #
4y“
4‰ œ (62.4) ˆ 2 156 24 ‰ œ
"
! (62.4)(11) 6
œ 114.4 lb
7. Using the coordinate system given in the accompanying figure, we see that the total width is L(y) œ 63 and the depth of the strip is (33.5 y) Ê F œ '0 w(33.5 y)L(y) dy 33
œ '0
33
64 1 #$
64 ‰ † (33.5 y) † 63 dy œ ˆ 12 (63)'0 (33.5 y) dy $ 33
$$ y# # “!
64 ‰ œ ˆ 12 (63) ’33.5y $
œ
(64)(63)(33)(67 33) (#) a12$ b
‰ ’(33.5)(33) œ ˆ 641#†63 $
33# # “
œ 1309 lb
8. (a) Use the coordinate system given in the accompanying ‰ figure. The depth of the strip is ˆ 11 6 y ft Ê F œ '0
11Î6
‰ w ˆ 11 6 y (width) dy
œ (62.4)(width)'0
11Î6
ˆ 11 ‰ 6 y dy
œ (62.4)(width) ’ 11 6 y
""Î' y# # “!
#
‰ † "# “ Ê Fend œ (62.4)(2) ˆ 121 ‰ ˆ "# ‰ ¸ 209.73 lb and Fside œ (62.4)(4) ˆ 121 ‰ ˆ "# ‰ ¸ 419.47 lb œ (62.4)(width) ’ˆ 11 6 36 36 (b) Use the coordinate system given in the accompanying figure. Find Y from the condition that the entire volume of the water is conserved (no spilling): 11 6 †2†4œ 2†2†Y 11 ‰ Ê Y œ 3 ft. The depth of a typical strip is ˆ 11 3 y ft and the total width is L(y) œ 2 ft. Thus, F œ '0
113
‰ w ˆ 11 3 y L(y) dy
11 ‰ œ '0 (62.4) ˆ 11 3 y † 2 dy œ (62.4)(2) ’ 3 y 113
""Î$ y# # “!
‰# “ œ œ (62.4)(2) ’ˆ "# ‰ ˆ 11 3
(62.4)(12") 9
force doubles.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
¸ 838.93 lb Ê the fluid
Section 6.7 Fluid Pressures and Forces 9. Using the coordinate system given in the accompanying figure, we see that the right-hand edge is x œ È1 y# so the total width is L(y) œ 2x œ 2È1 y# and the depth of the strip is (y). The force exerted by the water is therefore F œ 'c1 w † (y) † 2È1 y# dy 0
œ 62.4'c1 È1 y# d a1 y# b œ 62.4 ’ 23 a1 y# b 0
$Î# !
“
"
œ (62.4) ˆ 23 ‰ (1 0) œ 416 lb
10. Using the same coordinate system as in Exercise 15, the right-hand edge is x œ È3# y# and the total width is L(y) œ 2x œ 2È9 y# . The depth of the strip is (y). The force exerted by the milk is therefore F œ 'c3 w † (y) † 2È9 y# dy œ 64.5'c3 È9 y# d a9 y# b œ 64.5 ’ 23 a9 y# b 0
0
$Î# !
“
œ (64.5)(18) œ 1161 lb
$
œ (64.5) ˆ 23 ‰ (27 0)
11. The coordinate system is given in the text. The right-hand edge is x œ Èy and the total width is L(y) œ 2x œ 2Èy. (a) The depth of the strip is (2 y) so the force exerted by the liquid on the gate is F œ '0 w(2 y)L(y) dy 1
" œ '0 50(2 y) † 2Èy dy œ 100 '0 (2 y)Èy dy œ 100'0 ˆ2y"Î# y$Î# ‰ dy œ 100 43 y$Î# 25 y&Î# ‘ ! 1
1
1
‰ œ 100 ˆ 43 25 ‰ œ ˆ 100 15 (20 6) œ 93.33 lb
2‰ (b) We need to solve 160 œ '0 w(H y) † 2Èy dy for h. 160 œ 100 ˆ 2H 3 5 Ê H œ 3 ftÞ 1
12. Use the coordinate system given in the accompanying figure. The total width is L(y) œ 1. (a) The depth of the strip is (3 1) y œ (2 y) ft. The force exerted by the fluid in the window is F œ '0 w(2 y)L(y) dy œ 62.4 '0 (2 y) † 1 dy œ (62.4) ’2y 1
1
" y# # “!
œ (62.4) ˆ2 "# ‰ œ
(62.4)(3) #
œ 93.6 lb
(b) Suppose that H is the maximum height to which the tank can be filled without exceeding its design limitation. This means that the depth of a typical strip is (H 1) y and the force is F œ '0 w[(H 1) y]L(y) dy œ Fmax , where 1
Fmax œ 312 lb. Thus, Fmax œ w'0 [(H 1) y] † 1 dy œ (62.4) ’(H 1)y
" y# # “!
1
œ (62.4) ˆH 3# ‰
‰ (2H 3) œ 93.6 62.4H. Then Fmax œ 93.6 62.4H Ê 312 œ 93.6 62.4H Ê H œ œ ˆ 62.4 #
405.6 62.4
œ 6.5 ft
13. Suppose that h is the maximum height. Using the coordinate system given in the text, we find an equation for the line of the end plate's right-hand edge is y œ 5# x Ê x œ 25 y. The total width is L(y) œ 2x œ 45 y and the depth of the typical horizontal strip at level y is (h y). Then the force is F œ '0 w(h y)L(y) dy œ Fmax , h
where Fmax œ 6667 lb. Hence, Fmax œ w'0 (h y) † 45 y dy œ (62.4) ˆ 45 ‰ ' ahy y# b dy h
h
0
œ
# (62.4) ˆ 45 ‰ ’ hy#
h
y$ 3 “0
$
œ (62.4) ˆ 45 ‰ Š h#
$
h 3
‰ œ $Ɉ 54 ‰ ˆ 6667 10.4 ¸ 9.288 ft. The volume of water which the tank can hold is V œ Height œ h and
" #
(Base) œ
2 5
$
max ‰ ‹ œ (62.4) ˆ 45 ‰ ˆ "6 ‰ h$ œ (10.4) ˆ 45 ‰ h$ Ê h œ Ɉ 54 ‰ ˆ F10.4
" #
(Base)(Height) † 30, where
h Ê V œ ˆ 25 h# ‰ (30) œ 12h# ¸ 12(9.288)# ¸ 1035 ft$ .
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
409
410
Chapter 6 Applications of Definite Integrals
14. (a) After 9 hours of filling there are V œ 1000 † 9 œ 9000 cubic feet of water in the pool. The level of the water V is h œ Area , where Area œ 50 † 30 œ 1500 Ê h œ 9000 1500 œ 6 ft. The depth of the typical horizontal strip at level y is then (6 y) for the coordinate system given in the text. An equation for the drain plate's right-hand edge is y œ x Ê total width is L(y) œ 2x œ 2y. Thus the force against the drain plate is F œ '0 w(6 y)L(y) dy œ 62.4 '0 (6 y) † 2y dy œ (62.4)(2)'0 a6y y# b œ (62.4)(2) ’ 6y# 1
1
1
#
œ (124.8) ˆ3 "3 ‰ œ (124.8) ˆ 83 ‰ œ 332.8 lb
" y$ 3 “!
(b) Suppose that h is the maximum height. Then, the depth of a typical strip is (h y) and the force F œ '0 w(h y)L(y) dy œ Fmax , where Fmax œ 520 lb. Hence, Fmax œ (62.4)'0 (h y) † 2y dy 1
1
œ 124.8'0 ahy y# b dy œ (124.8) ’ hy# 1
Êhœ
27 3
#
" y$ 3 “!
œ (124.8) ˆ h# "3 ‰ œ (20.8)(3h 2) Ê
520 20.8
œ 3h 2
œ 9 ft
15. The pressure at level y is p(y) œ w † y Ê the average pressure is p œ œ
# ˆ wb ‰ Š b# ‹
œ
" b
'0b p(y) dy œ b" '0b w † y dy œ b" w ’ y# “ b #
0
wb #
. This is the pressure at level
b #
, which
is the pressure at the middle of the plate. 16. The force exerted by the fluid is F œ '0 w(depth)(length) dy œ '0 w † y † a dy œ (w † a)'0 y dy œ (w † a) ’ y# “ b
œ
# w Š ab# ‹
œ
ˆ wb ‰ # (ab)
b
b
#
b 0
œ p † Area, where p is the average value of the pressure (see Exercise 21).
17. When the water reaches the top of the tank the force on the movable side is 'c2 (62.4) ˆ2È4 y# ‰ (y) dy 0
œ (62.4)'c2 a4 y# b 0
"Î#
(2y) dy œ (62.4) ’ 23 a4 y# b
$Î# !
“
#
œ (62.4) ˆ 23 ‰ ˆ4$Î# ‰ œ 332.8 ft † lb. The force
compressing the spring is F œ 100x, so when the tank is full we have 332.8 œ 100x Ê x ¸ 3.33 ft. Therefore the movable end does not reach the required 5 ft to allow drainage Ê the tank will overflow. 18. (a) Using the given coordinate system we see that the total width is L(y) œ 3 and the depth of the strip is (3 y).
Thus, F œ '0 w(3 y)L(y) dy œ '0 (62.4)(3 y) † 3 dy 3
3
œ (62.4)(3)'0 (3 y) dy œ (62.4)(3) ’3y 3
$ y# # “!
œ (62.4)(3) ˆ9 9# ‰ œ (62.4)(3) ˆ 9# ‰ œ 842.4 lb
(b) Find a new water level Y such that FY œ (0.75)(842.4 lb) œ 631.8 lb. The new depth of the strip is (Y y) and Y is the new upper limit of integration. Thus, FY œ '0 w(Y y)L(y) dy Y
œ 62.4'0 (Y y) † 3 dy œ (62.4)(3)'0 (Y y) dy œ (62.4)(3) ’Yy Y
Y
Y
y# # “0
œ (62.4)(3) ŠY#
Y# # ‹
#
2FY È6.75 ¸ 2.598 ft. So, ?Y œ 3 Y œ (62.4)(3) Š Y# ‹ . Therefore, Y œ É (62.4)(3) œ É 1263.6 187.2 œ
¸ 3 2.598 ¸ 0.402 ft ¸ 4.8 in 19. Use a coordinate system with y œ 0 at the bottom of the carton and with L(y) œ 3.75 and the depth of a typical strip being (7.75 y). Then F œ '0
7Þ75
' ‰ w(7.75 y)L(y) dy œ ˆ 64.5 12$ (3.75) 0
7Þ75
‰ (7.75 y) dy œ ˆ 64.5 12$ (3.75) ’7.75y
#
(7.75) ‰ œ ˆ 64.5 ¸ 4.2 lb 12$ (3.75) #
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
(Þ(& y# # “!
Chapter 6 Practice Exercises
411
57 ‰ 20. The force against the base is Fbase œ pA œ whA œ w † h † (length)(width) œ ˆ 12 (10)(5.75)(3.5) ¸ 6.64 lb. $
To find the fluid force against each side, use a coordinate system with y œ 0 at the bottom of the can, so that the depth of a of ‰ of ‰ ˆ 57 ‰ ˆ width typical strip is (10 y): F œ '0 w(10 y) ˆ width the side dy œ 12$ the side ’10y 10
"! y# # “!
57 ‰ ˆ width of ‰ ˆ 100 ‰ 57 ‰ 57 ‰ œ ˆ 12 Ê Fend œ ˆ 12 (50)(3.5) ¸ 5.773 lb and Fside œ ˆ 12 (50)(5.75) ¸ 9.484 lb $ $ $ the side #
21. (a) An equation of the right-hand edge is y œ
x Ê xœ
3 #
2 3
y and L(y) œ 2x œ
4y 3
. The depth of the strip
is (3 y) Ê F œ '0 w(3 y)L(y) dy œ '0 (62.4)(3 y) ˆ 43 y‰ dy œ (62.4) † ˆ 43 ‰'0 a3y y# b dy 3
3
$ y$ 3 “!
œ (62.4) ˆ 43 ‰ ’ 3# y#
œ (62.4) ˆ 43 ‰ 27 #
3
27 ‘ 3
‰ œ (62.4) ˆ 34 ‰ ˆ 27 6 œ 374.4 lb
(b) We want to find a new water level Y such that FY œ
" #
(374.4) œ 187.2 lb. The new depth of the strip is
(Y y), and Y is the new upper limit of integration. Thus, FY œ '0 w(Y y)L(y) dy Y
œ 62.4'0 (Y y) ˆ 43 y‰ dy œ (62.4) ˆ 43 ‰'0 aYy y# b dy œ (62.4) ˆ 43 ‰ ’Y † Y
Y
œ (62.4) ˆ 29 ‰ Y$ . Therefore Y$ œ
9FY 2†(62.4)
œ
(9)(187.2) 124.8
y# #
Y
y$ 3 “!
$
œ (62.4) ˆ 34 ‰ Š Y2
Y$ 3 ‹
$ $È Ê Y œ É (9)(187.2) 13.5 ¸ 2.3811 ft. So, 124.8 œ
?Y œ 3 Y ¸ 3 2.3811 ¸ 0.6189 ft ¸ 7.5 in. to the nearest half inch. (c) No, it does not matter how long the trough is. The fluid pressure and the resulting force depend only on depth of the water. ‰ 22. The area of a strip of the face of height ?y and parallel to the base is 100ˆ 26 24 † ?y, where the factor of
26 24
inclination of the face of the dam. With the origin at the bottom of the dam, the force on the face is then: ‰ F œ '0 w(24 y)a100bˆ 26 24 dy œ '('! ’#%y 24
#%
y# # “!
œ '('!Š#%#
#%# # ‹
œ 1,946,880 lb.
CHAPTER 6 PRACTICE EXERCISES # 1. A(x) œ 14 (diameter)# œ 14 ˆÈx x# ‰ œ 14 ˆx 2Èx † x# x% ‰ ; a œ 0, b œ 1
Ê V œ 'a A(x) dx œ b
œ
1 4
œ
1 4†70
#
’ x# 74 x(Î#
È3 4
x& 5 “!
(35 40 14) œ
2. A(x) œ œ
1 4 "
" #
'01 ˆx 2x&Î# x% ‰ dx 1 4
œ
ˆ "#
4 7
5" ‰
91 280
(side)# ˆsin 13 ‰ œ
È3 4
ˆ2Èx x‰#
ˆ4x 4xÈx x# ‰ ; a œ 0, b œ 4
Ê V œ 'a A(x) dx œ
È3 4
b
œ
È3 4
œ
32È3 4
’2x# 85 x&Î# ˆ1
8 5
'04 ˆ4x 4x$Î# x# ‰ dx
% x$ 3 “!
32 ‰ œ
8È 3 15
œ
È3 4
ˆ32
8†32 5
(15 24 10) œ
64 ‰ 3
8È 3 15
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
accounts for the
412
Chapter 6 Applications of Definite Integrals
3. A(x) œ œ
1 4
1 4
(diameter)# œ
1 4
(2 sin x 2 cos x)#
† 4 asin# x 2 sin x cos x cos# xb
œ 1(1 sin 2x); a œ
1 4
,bœ
51 4
Ê V œ 'a A(x) dx œ 1 '1Î4 (1 sin 2x) dx 51Î4
b
œ 1 x
cos 2x ‘ &1Î% # 1Î%
œ 1 ’Š 541
cos 5#1 #
cos 1# #
‹ Š 14
‹“ œ 1 # #
#
%
4. A(x) œ (edge)# œ ŒŠÈ6 Èx‹ 0 œ ŠÈ6 Èx‹ œ 36 24È6 Èx 36x 4È6 x$Î# x# ; a œ 0, b œ 6 Ê V œ 'a A(x) dx œ '0 Š36 24È6 Èx 36x 4È6 x$Î# x# ‹ dx b
6
œ ’36x 24È6 † 23 x$Î# 18x# 4È6 † 25 x&Î# œ 216 576 648 5. A(x) œ
(diameter)# œ
1 4
72 œ 360
Š2Èx
x# 4‹
#
œ
1728 5
œ
1 4
œ 216 16 † È6 È6 † 6 18 † 6# 58 È6 È6 † 6#
18001728 5
Š4x x&Î#
œ
6$ 3
72 5
x% 16 ‹ ;
a œ 0, b œ 4 Ê V œ 'a A(x) dx b
'04 Š4x x&Î# 16x ‹ dx œ 14 ’2x# 27 x(Î# 5x†16 “ % œ 14 ˆ32 32 † 87 25 † 32‰ %
œ
1 4
œ
321 4
ˆ1
6. A(x) œ œ
1 4
1728 5
' x$ 3 “!
È3 4
" #
8 7
25 ‰ œ
81 35
&
(35 40 14) œ
(edge)# sin ˆ 13 ‰ œ
È3 4
!
721 35
2Èx ˆ2Èx‰‘#
ˆ4Èx‰# œ 4È3 x; a œ 0, b œ 1
Ê V œ 'a A(x) dx œ '0 4È3 x dx œ ’2È3 x# “ b
1
" !
œ 2È3
7. (a) .3=5 7/>29. :
V œ 'a 1R# (x) dx œ 'c1 1 a3x% b dx œ 1 'c1 9x) dx b
1
1
#
"
œ 1 cx* d " œ 21
(b) =2/66 7/>29. : shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x a3x% b dx œ 21 † 3'0 x& dx œ 21 † 3 ’ x6 “ œ 1 b
1
1
'
!
Note: The lower limit of integration is 0 rather than 1. (c) =2/66 7/>29. : shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ 21'c1 (1 x) a3x% b dx œ 21 ’ 3x5 b
1
"
&
(d) A+=2/< 7/>29. :
" x' 2 “ "
œ 21 ˆ 35 "# ‰ ˆ 35 "# ‰‘ œ
R(x) œ 3, r(x) œ 3 3x% œ 3 a1 x% b Ê V œ 'a 1 cR# (x) r# (x)d dx œ 'c1 1 ’9 9 a1 x% b “ dx b
1
œ 91 'c1 c1 a1 2x% x) bd dx œ 91 'c1 a2x% x) b dx œ 91 ’ 2x5 1
1
&
" x* 9 “ "
#
œ 181 25 9" ‘ œ
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
21†13 5
œ
261 5
121 5
Chapter 6 Practice Exercises 8. (a) A+=2/< 7/>29. : R(x) œ
, r(x) œ
4 x$
" #
# # # & Ê V œ 'a 1cR# (x) r# (x)d dx œ '1 1 ’ˆ x4$ ‰ ˆ "# ‰ “ dx œ 1 16 x4 ‘ " 5 x b
2
"‰ " ˆ 16 " ‰‘ œ 1 ˆ 10 œ 1 ˆ 5†16 32 # 5 4
(b) =2/66 7/>29. :
V œ 21'1 x ˆ x4$ "# ‰ dx œ 21 ’4x" 2
(c) =2/66 7/>29. :
" #
# x# 4 “"
4" ‰ œ
16 5
1 20
(2 10 64 5) œ
b
2
4 x
x
(d) A+=2/< 7/>29. :
# x# 4 “"
571 #0
œ 21 ˆ 4# 1‰ ˆ4 4" ‰‘ œ 21 ˆ 54 ‰ œ
shell ‰ shell V œ 21'a ˆ radius Š height ‹ dx œ 21'1 (2 x) ˆ x4$ "# ‰ dx œ 21'1 ˆ x8$
œ 21 ’ x4#
413
2
4 x#
51 #
1 x# ‰ dx
œ 21 (1 2 2 1) ˆ4 4 1 4" ‰‘ œ
31 #
V œ 'a 1cR# (x) r# (x)d dx b
# œ 1 '1 ’ˆ 7# ‰ ˆ4 2
dx
œ
491 4
161'1 a1 2x$ x' b dx
œ
491 4
161 ’x x#
œ
491 4 491 4 491 4
161 ˆ2 4" 5†"3# ‰ ˆ1 1 5" ‰‘ " 161 ˆ 4" 160 5" ‰
œ œ 9.
4 ‰# x$ “
2
161 160
# x& 5 “"
(40 1 32) œ
491 4
711 10
1031 20
œ
(a) .3=5 7/>29. :
V œ 1 '1 ŠÈx 1‹ dx œ 1'1 (x 1) dx œ 1 ’ x# x“ #
5
5
#
‰ ˆ" ‰‘ œ 1 ˆ 24 ‰ œ 1 ˆ 25 # 5 # 1 # 4 œ 81
& "
(b) A+=2/< 7/>29. :
R(y) œ 5, r(y) œ y# 1 Ê V œ 'c 1 cR# (y) r# (y)d dy œ 1 'c2 ’25 ay# 1b “ dy d
2
œ 1'c2 a25 y% 2y# 1b dy œ 1 'c2 a24 y% 2y# b dy œ 1 ’24y 2
2
œ 321 ˆ3
2 5
"3 ‰ œ
321 15
(45 6 5) œ
10881 15
#
y& 5
23 y$ “
# #
œ 21 ˆ24 † 2
(c) .3=5 7/>29. : R(y) œ 5 ay# 1b œ 4 y#
Ê V œ 'c 1R# (y) dy œ 'c2 1 a4 y# b dy d
2
#
œ 1 'c2 a16 8y# y% b dy 2
œ 1 ’16y
8y$ 3
œ 641 ˆ1
2 3
# y& 5 “ #
"5 ‰ œ
œ 21 ˆ32
641 15
64 3
(15 10 3) œ
32 ‰ 5 5121 15
10. (a) =2/66 7/>29. :
shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21y Šy d
4
œ 21'0 Šy# 4
œ
21 1#
† 64 œ
y$ 4‹
321 3
$
dy œ 21 ’ y3
%
y% 16 “ !
y# 4‹
dy
œ 21 ˆ 64 3
64 ‰ 4
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
32 5
2 3
† 8‰
414
Chapter 6 Applications of Definite Integrals
(b) =2/66 7/>29. : shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x ˆ2Èx x‰ dx œ 21'0 ˆ2x$Î# x# ‰ dx œ 21 ’ 45 x&Î# b
4
œ 21 ˆ 45 † 32
64 ‰ 3
œ
4
1281 15
% x$ 3 “!
(c) =2/66 7/>29. :
shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21(4 x) ˆ2Èx x‰ dx œ 21'0 ˆ8x"Î# 4x 2x$Î# x# ‰ dx b
4
$Î# œ 21 ’ 16 2x# 54 x&Î# 3 x
œ 641 ˆ1 45 ‰ œ
641 5
4
% x$ 3 “!
œ 21 ˆ 16 3 † 8 32
(d) =2/66 7/>29. :
shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21(4 y) Šy d
4
œ 21'0 Š4y 2y# 4
y$ 4‹
y# 4‹ %
y% 16 “ !
dy œ 21 ’2y# 23 y$
4 5
† 32
64 ‰ 3
œ 641 ˆ 34 1
dy œ 21'0 Š4y y# y# 4
œ 21 ˆ32
2 3
4 5
y$ 4‹
32 ‰
dy
† 64 16‰ œ 321 ˆ2
321 3
1‰ œ
8 3
11. .3=5 7/>29. : R(x) œ tan x, a œ 0, b œ
1 3
Ê V œ 1 '0 tan# x dx œ 1'0 asec# x 1b dx œ 1[tan x x]! 1Î3
12. .3=5 7/>29. :
1Î3
1Î$
V œ 1'0 (2 sin x)# dx œ 1 '0 a4 4 sin x sin# xb dx œ 1'0 ˆ4 4 sin x 1
œ 1 4x 4 cos x
1
x #
sin 2x ‘ 1 4 !
1
œ 1 ˆ41 4
1 #
0‰ (0 4 0 0)‘ œ
œ
1cos 2x ‰ dx # 9 1 1 ˆ # 8‰ œ 1#
1 Š3È31‹ 3
(91 16)
13. (a) .3=5 7/>29. :
V œ 1'0 ax# 2xb dx œ 1'0 ax% 4x$ 4x# b dx œ 1 ’ x5 x% 43 x$ “ œ 1 ˆ 32 5 16 2
œ
161 15
2
#
(6 15 10) œ
#
&
!
161 15
32 ‰ 3
(b) A+=2/< 7/>29. :
V œ '0 1’1# ax# 2x "b “ dx œ '0 1 dx '0 1 ax "b% dx œ #1 ’1 2
2
#
2
(c) =2/66 7/>29. :
# ax"b& & “!
œ #1 1 †
# &
œ
)1 &
shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ 21'0 (2 x) c ax# 2xbd dx œ 21'0 (2 x) a2x x# b dx b
2
2
œ 21'0 a4x 2x# 2x# x$ b dx œ 21'0 ax$ 4x# 4xb dx œ 21 ’ x4 43 x$ 2x# “ œ 21 ˆ4 2
œ
21 3
2
(36 32) œ
#
%
!
81 3
32 3
8‰
(d) A+=2/< 7/>29. :
V œ 1 '0 c2 ax# 2xbd dx 1'0 2# dx œ 1'0 ’4 4 ax# 2xb ax# 2xb “ dx 81 2
2
#
2
#
œ 1'0 a4 4x# 8x x% 4x$ 4x# b dx 81 œ 1'0 ax% 4x$ 8x 4b dx 81 2
&
2
#
‰ œ 1 ’ x5 x% 4x# 4x“ 81 œ 1 ˆ 32 5 16 16 8 81 œ !
1 5
(32 40) 81 œ
721 5
14. .3=5 7/>29. :
V œ 21'0 4 tan# x dx œ 81'0 asec# x 1b dx œ 81[tan x x]! 1Î4
1Î4
1Î%
œ 21(4 1)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
401 5
œ
321 5
Chapter 6 Practice Exercises 15. The material removed from the sphere consists of a cylinder and two "caps." From the diagram, the height of the cylinder #
is 2h, where h# ŠÈ$‹ œ ## , i.e. h œ ". Thus #
Vcyl œ a#hb1ŠÈ$‹ œ '1 ft$ . To get the volume of a cap, use the disk method and x# y# œ ## : Vcap œ '" 1x# dy 2
œ '" 1a% y# bdy œ 1’%y 2
œ 1ˆ8 83 ‰ ˆ% "3 ‰‘ œ
# y3 3 “"
&1 3
ft$ . Therefore,
Vremoved œ Vcyl #Vcap œ '1
"!1 3
#)1 3
œ
ft$ .
16. We rotate the region enclosed by the curve y œ É12 ˆ1
4x# ‰ 121
and the x-axis around the x-axis. To find the
11Î2
volume we use the .3=5 method: V œ 'a 1R# (x) dx œ 'c11Î2 1 ŠÉ12 ˆ1 b
œ 121'c11Î2 Š1 11Î2
4x# 121 ‹
œ 1321 ˆ1 "3 ‰ œ 17. y œ x"Î#
x$Î# 3
Ê
ˆ4
" #
œ
#
x"Î# "# x"Î# Ê Š dy dx ‹ œ
" 4
œ '1
8
dx dy
È9x#Î$ 4 3x"Î$
œ
5 12
2 3
#
4
4
ˆ2
14 ‰ 3
#
x"Î$ Ê Š dx dy ‹ œ
œ
4x#Î$ 9
" #
ˆx"Î# x"Î# ‰ dx œ
" #
2x"Î# 23 x$Î# ‘ % "
10 3 dx Ê L œ '1 Ê1 Š dy ‹ dy œ '1 É1 #
8
8
4 9x#Î$
dy
'1 È9x#Î$ 4 ˆx"Î$ ‰ dx; u œ 9x#Î$ 4 Ê du œ 6y"Î$ dy; x œ 1 40 " ' " 2 $Î# ‘ %! Ä L œ 18 u"Î# du œ 18 œ #"7 40$Î# 13$Î# ‘ ¸ 7.634 3 u "$ 13
dx œ
x œ 8 Ê u œ 40d 19. y œ
" #
† 8‰ ˆ2 23 ‰‘ œ
18. x œ y#Î$ Ê
8
" 3
x'Î& 58 x%Î& Ê
dx
ˆ x" 2 x‰ Ê L œ ' É1 4" ˆ x" 2 x‰ dx 1
4
2 3
4x# 121 ‹
œ 881 ¸ 276 in$
4
" #
11Î2
dx œ 1 '11Î2 12 Š1
4 ‰ 11 ˆ 4 ‰ ˆ 11 ‰ “ œ 1321 ’1 ˆ 363 œ 241 ’ 11 Š 4 ‹“ 2 363 #
# Ê L œ '1 É 4" ˆ x" 2 x‰ dx œ '1 É 4" ax"Î# x"Î# b dx œ '1
œ
#
$
4x$ 363 “ ""Î#
dx œ 121 ’x
2641 3
dy dx
""Î#
4x# ‰ 121 ‹
#
" #
œ
dy dx
x"Î& "# x"Î& Ê Š dy dx ‹ œ
" 4
Ê u œ 13,
ˆx#Î& 2 x#Î& ‰
# Ê L œ '1 É1 4" ax#Î& 2 x#Î& b dx Ê L œ '1 É 4" ax#Î& 2 x#Î& b dx œ ' É 4" ax"Î& x"Î& b dx 32
32
32
1
œ '1
32
œ
" 48
20. x œ
" #
ˆx
"Î&
x
"Î& ‰
(1260 450) œ
" 1#
y$
" y
Ê
" % œ '1 É 16 y 2
" #
dx œ œ
1710 48
dx dt
5 4
$# x%Î& ‘ "
#
œ
" 4
" y%
dy œ '1 ÊŠ 4" y#
y#
" y#
x
'Î&
dx dy
8 œ ˆ 12 "# ‰ ˆ 1"# 1‰ œ
21.
" 5 # 6 285 8
dx Ê Š dy ‹ œ
2
7 1#
œ 5 sin t 5 sin 5t and
dy dt
" #
œ
" y# ‹
" 16 #
œ
" #
ˆ 65
y%
" #
'
†2
" y%
5 4
ˆ 56
5 ‰‘ 4
œ
" #
ˆ 315 6
" % Ê L œ '1 Ê1 Š 16 y
dy œ '1 Š 4" y# 2
†2
%‰
2
" y# ‹
dy œ ’ 1"# y$ y" “
" #
75 ‰ 4
" y% ‹
dy
# "
13 12
#
‰ Š dy œ 5 cos t 5 cos 5t Ê Êˆ dx dt dt ‹
#
œ Éa5 sin t 5 sin 5tb# a5 cos t 5 cos 5tb# œ 5Èsin# 5t #sin t sin 5t sin# t cos# t #cos t cos 5t cos# 5t œ &È# #asin t sin 5t cos t cos 5 tb œ 5È#a" cos %tb œ 5É%ˆ "# ‰a" cos %tb œ "!Èsin# #t œ "!lsin #tl œ "!sin #t (since ! Ÿ t Ÿ 1# ) Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
415
416
Chapter 6 Applications of Definite Integrals Ê Length œ '!
1 Î2
22.
dx dt
œ 3t2 12t and
1Î#
"!sin #t dt œ c5 cos #td ! dy dt
œ a&ba"b a&ba"b œ "! #
#
‰ Š dy Éa3t2 12tb# a3t2 12tb# œ È288t# "8t4 œ 3t2 12t Ê Êˆ dx dt dt ‹ œ
œ 3È2 ktkÈ16 t2 Ê Length œ '! 3È2 ktkÈ16 t2 dt œ 3È2'! t È16 t2 dt; ’u œ 16 t2 Ê du œ 2t dt "
"
3È 2 2
Ê "# du œ t dt; t œ 0 Ê u œ 16; t œ 1 Ê u œ 17“; œ
23.
dx d)
3È 2 2
† 23 Ša17b3/2 64‹ œ È2Ša17b3/2 64‹ ¸ 8.617.
œ $ sin ) and
Ê Length œ '!
dy d)
$1Î2
#
24. x œ t and y œ œ'
'16"7 Èu du œ 3È2 2 23 u3/2 ‘1617 œ 3È2 2 Š 23 a17b3/2 23 a16b3/2 ‹
t$ 3
$ d) œ $'!
$1Î2
d) œ $ˆ $#1 !‰ œ
t, È3 Ÿ t Ÿ È3 Ê
È3
È 3
#
#
‰ Š dy Éa$ sin )b# a$ cos )b# œ È$asin# ) cos# )b œ $ œ $ cos ) Ê Êˆ dx d) d) ‹ œ
Èt% #t# " dt œ
'
dx dt
*1 #
œ 2t and
dy dt
È3
È 3
Èt% 2t# " dt œ
œt
'
#
" Ê Length œ '
È3
È 3
È3
È 3
Éat# "b# dt œ
Éa2tb# at# "b# dt
È
'È33 at# "b dt œ ’ t3 t“ 3
È3 È 3
œ 4È3 25. Intersection points: 3 x# œ 2x# Ê 3x# 3 œ 0 Ê 3(x 1)(x 1) œ 0 Ê x œ 1 or x œ 1. Symmetry suggests that x œ 0. The typical @/3-+6 strip has # # # center of mass: (µ x ßµ y ) œ Šxß 2x a3 x b ‹ œ Šxß x 3 ‹ , #
#
#
#
#
length: a3 x b 2x œ 3 a1 x b, width: dx, area: dA œ 3 a1 x# b dx, and mass: dm œ $ † dA œ 3$ a1 x# b dx Ê the moment about the x-axis is µ y dm œ œ
3 #
3 #
$ ax# 3b a1 x# b dx œ &
$ ’ x5
œ 3$ ’x
2x$ 3
" x$ 3 “ "
3x“
" "
3 #
$ ax% 2x# 3b dx Ê Mx œ ' µ y dm œ
œ 3$ ˆ 5"
2 3
3‰ œ
œ 6$ ˆ1 "3 ‰ œ 4$ Ê y œ
Mx M
œ
3$ 15
(3 10 45) œ
32$ 5 †4 $
œ
8 5
32$ 5
3 #
$ 'c1 ax% 2x# 3b dx "
; M œ ' dm œ 3$ 'c1 a1 x# b dx "
. Therefore, the centroid is (xß y) œ ˆ!ß 85 ‰ .
26. Symmetry suggests that x œ 0. The typical @/3-+6 # strip has center of mass: (µ x ßµ y ) œ Šxß x# ‹ , length: x# , width: dx, area: dA œ x# dx, mass: dm œ $ † dA œ $ x# dx Ê the moment about the x-axis is µ y dm œ #$ x# † x# dx x% dx Ê Mx œ ' µ y dm œ
œ
$ #
œ
2$ 10
a2& b œ
32$ 5
$ #
; M œ ' dm œ $
'c22 x% dx œ 10$ cx& d ## 'c22 x# dx œ $ ’ x3 “ # $
#
œ
2$ 3
a2$ b œ
16$ 3
Ê yœ
Mx M
œ
32†$ †3 5†16†$
œ
centroid is(xß y) œ ˆ!ß 65 ‰ .
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
6 5
. Therefore, the
Chapter 6 Practice Exercises
417
27. The typical @/3-+6 strip has: center of mass: (µ x ßµ y ) œ Œxß
#
4 x4 #
, length: 4
area: dA œ Š4 œ $ Š4
x# 4‹
x# 4 ‹dx,
width: dx,
mass: dm œ $ † dA
dx Ê the moment about the x-axis is #
Š4 x4 ‹
µ y dm œ $ †
x# 4,
x# 4‹
Š4
#
$ #
dx œ
moment about the y-axis is µ x dm œ $ Š4 œ
$ 2
’16x
% x& 5†16 “ !
$ #
œ
64
#
œ
16†$ †3 32†$
œ
Mx M
and y œ
3 2
œ
dx. Thus, Mx œ ' µ y dm œ
4
4
My M
x$ 4‹
‹ † x dx œ $ Š4x
x 4
œ $ (32 16) œ 16$ ; M œ ' dm œ $ '0 Š4 Ê xœ
dx; the
; My œ ' µ x dm œ $ '0 Š4x
128$ 5
œ
64 ‘ 5
x% 16 ‹
Š16
128†$ †3 5†32†$
x# 4‹
œ
dx œ $ ’4x
12 5
% x$ 12 “ !
x$ 4‹
dx œ $ ’2x#
œ $ ˆ16
64 ‰ 1#
$ #
'04 Š16 16x ‹ dx %
% x% 16 “ !
32$ 3
œ
‰ . Therefore, the centroid is (xß y) œ ˆ 3# ß 12 5 .
28. A typical 29+6 strip has: # center of mass: (µ x ßµ y ) œ Š y # 2y ß y‹ , length: 2y y# , width: dy, area: dA œ a2y y# b dy, mass: dm œ $ † dA œ $ a2y y# b dy; the moment about the x-axis is µ y dm œ $ † y † a2y y# b dy œ $ a2y# y$ b ; the moment # about the y-axis is µ x dm œ $ † ay 2yb † a2y y# b dy #
œ a4y y b dy Ê Mx œ ' µ y dm œ $ '0 a2y# y$ b dy $ #
#
œ $ ’ 23 y$ œ
$ #
ˆ 43†8
yœ
Mx M
2
%
œ
# y% 4 “!
32 ‰ 5
4†$ †3 3†4†$
œ
œ $ ˆ 23 † 8 32$ 15
16 ‰ 4
œ $ ˆ 16 3
16 ‰ 4
œ
$ †16 12
œ
4$ 3
; My œ ' µ x dm œ
$ #
'02 a4y# y% b dy œ #$ ’ 34 y$ y5 “ # &
$ ; M œ ' dm œ $ '0 a2y y# b dy œ $ ’y# y3 “ œ $ ˆ4 83 ‰ œ
#
2
!
!
4$ 3
Ê xœ
My M
œ
$ †32†3 15†$ †4
œ
œ 1. Therefore, the centroid is (xß y) œ ˆ 85 ß 1‰ .
29. A typical horizontal strip has: center of mass: (µ x ßµ y ) œ Šy
#
2y # ß y‹ ,
length: 2y y# , width: dy,
area: dA œ a2y y# b dy, mass: dm œ $ † dA œ (1 y) a2y y# b dy Ê the moment about the x-axis is µ y dm œ y(1 y) a2y y# b dy # œ a2y 2y$ y$ y% b dy œ a2y# y$ y% b dy; the moment about the y-axis is # µ x dm œ Š y 2y ‹ (1 y) a2y y# b dy œ " a4y# y% b (1 y) dy œ #
Ê Mx œ ' µ y dm œ '0 a2y# y$ y% b dy œ ’ 23 y$ 2
œ
16 60
œ
" #
" #
#
(20 15 24) œ $
Š 4†32 2%
2& 5
4 15
(11) œ
2' 6‹
‰ ˆ 38 ‰ œ œ ˆ 44 15
44 40
œ
11 10
y$ 3
; My œ ' µ x dm œ '0
2
œ 4 ˆ 43 2
œ '0 a2y y# y$ b dy œ ’y# 2
44 15
y% 4
4 5
" #
# y& 5 “!
œ ˆ4
8 3
16 ‰ 4
œ ˆ 16 3
œ
8 3
24 5
area: dA œ
x$Î#
32 ‰ 5
œ 16 ˆ "3 " #
dx, mass: dm œ $ † dA œ $ †
x$Î#
25 ‰
’ 43 y$ y%
y& 5
2
Ê xœ
My M
‰ ˆ 83 ‰ œ œ ˆ 24 5
9 5
‰ . Therefore, the center of mass is (xß y) œ ˆ 95 ß 11 10 .
3
" 4
; M œ ' dm œ '0 (1 y) a2y y# b dy
30. A typical vertical strip has: center of mass: (µ x ßµ y ) œ ˆxß 2x3$Î# ‰ , length: 3
16 4
a4y# 4y$ y% y& b dy œ
86 ‰ œ 4 ˆ2 45 ‰ œ
# y% 4 “!
a4y# 4y$ y% y& b dy
3 x$Î#
, width: dx,
dx Ê the moment about the x-axis is
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
and y œ
Mx M
# y' 6 “!
8 5
and
418
Chapter 6 Applications of Definite Integrals
µ y dm œ
9$ 2x$
3 † $ x$Î# dx œ
3 #x$Î#
(a) Mx œ $ '1
9
M œ $ '1
9
(b) Mx œ '1
9
" #
9$ #
ˆ x9$ ‰ dx œ
#
*
20$ 9
’ x# “ œ "
*
ˆ x9$ ‰ dx œ
9 #
*
dx.
*
My M
œ
12$ 4$
œ 3 and y œ
Mx M
œ
ˆ 209$ ‰ 4$
œ
5 9
* 3 ‰ 3 ‰ "x ‘ * œ 4; My œ ' x# ˆ $Î# dx œ 2x$Î# ‘ " œ 52; M œ '1 x ˆ x$Î# dx " x 1 9
œ 6 x"Î# ‘ " œ 12 Ê x œ 31. S œ 'a 21y Ê1 Š dy dx ‹ dx; #
b
3$ x"Î#
3 ‰ ; My œ $ '1 x ˆ x$Î# dx œ 3$ 2x"Î# ‘ " œ 12$ ; 9
dx œ 6$ x"Î# ‘ " œ 4$ Ê x œ
3 x$Î# x #
3 dx; the moment about the y-axis is µ x dm œ x † $ x$Î# dx œ
My M
dy dx
œ
œ
13 3
and y œ
Mx M
9
œ
" 3
#
" È2x 1
Ê Š dy dx ‹ œ
Ê S œ '0 21È2x 1 É1 3
" #x 1
" #x 1
dx
2 È ' Èx 1 dx œ 2È21 2 (x 1)$Î# ‘ $ œ 2È21 † 2 (8 1) œ œ 21'0 È2x 1 É 2x 2x1 dx œ 2 21 0 3 3 ! 3
3
32. S œ 'a 21y Ê1 Š dy dx ‹ dx; #
b
œ
1 6
dy dx
% ' œ x# Ê Š dy dx ‹ œ x Ê S œ 0 21 † #
1
x$ 3
È1 x% dx œ
1 6
281È2 3
'01 È1 x% a4x$ b dx
'01 È1 x% d a1 x% b œ 16 ’ 32 a1 x% b$Î# “ " œ 19 ’2È2 1“ !
33. S œ 'c 21x Ê1 Š dx dy ‹ dy; #
d
dx dy
œ
ˆ "# ‰ (4 2y) È4y y#
2y È4y y#
œ
#
Ê 1 Š dx dy ‹ œ
4y y# 4 4y y# 4y y#
œ
4 4y y#
Ê S œ '1 21 È4y y# É 4y 4 y# dy œ 41'1 dx œ 41 2
2
34. S œ 'c 21x Ê1 Š dx dy ‹ dy; #
d
œ 1'2 È4y 1 dy œ 6
35. x œ
t# #
1 4
dx dy
œ
*
36. x œ t#
" 2t
761 3
" È2
1
œ 21 Š2
(125 27) œ
œ t and
ŸtŸ1 Ê
Ê Surface Area œ '1ÎÈ2 21 ˆt# 1
dx dt
1 6
dy dt
1 6
" 4y
œ
Ê S œ '2 21Èy † 6
4y 1 4y
(98) œ
È4y 1 È4y
dy
491 3
È5
œ 2 Ê Surface Area œ '0 21(2t)Èt# 4 dt œ '4 21u"Î# du 9
, where u œ t# 4 Ê du œ 2t dt; t œ 0 Ê u œ 4, t œ È5 Ê u œ 9
and y œ 4Èt ,
œ 21 '1ÎÈ2 ˆt#
#
Ê 1 Š dx dy ‹ œ 1
23 (4y 1)$Î# ‘ ' œ #
and y œ 2t, 0 Ÿ t Ÿ È5 Ê
œ 21 23 u$Î# ‘ % œ
1 2È y
" ‰ˆ 2t 2t
" ‰ 2t#
"‰ #t
dx dt
œ 2t
ʈ2t
" ‰# 2t#
dt œ 21 '1ÎÈ2 ˆ2t$ 1
" 2t#
and
dy dt
œ
2 Èt
Š È2 t ‹ dt œ 21 '1ÎÈ2 ˆt# #
3 #
1
" ‰ Ɉ 2t #t
" ‰# #t#
dt
"
4" t$ ‰ dt œ 21 2" t% 3# t 8" t# ‘ "ÎÈ#
3È 2 4 ‹
37. The equipment alone: the force required to lift the equipment is equal to its weight Ê F" (x) œ 100 N.
The work done is W" œ 'a F" (x) dx œ '0 100 dx œ [100x]%! ! œ 4000 J; the rope alone: the force required b
40
to lift the rope is equal to the weight of the rope paid out at elevation x Ê F# (x) œ 0.8(40 x). The work done is W# œ 'a F# (x) dx œ '0 0.8(40 x) dx œ 0.8 ’40x b
40
the total work is W œ W" W# œ 4000 640 œ 4640 J
%! x# # “!
œ 0.8 Š40#
40# # ‹
œ
(0.8)(1600) #
œ 640 J;
38. The force required to lift the water is equal to the water's weight, which varies steadily from 8 † 800 lb to 8 † 400 lb over the 4750 ft elevation. When the truck is x ft off the base of Mt. Washington, the water weight is x‰ F(x) œ 8 † 800 † ˆ 2†24750 œ (6400) ˆ1 †4750
x ‰ 9500
lb. The work done is W œ 'a F(x) dx b
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 6 Practice Exercises œ '0
4750
6400 ˆ1
x ‰ 9500
dx œ 6400 ’x
œ 22,800,000 ft † lb
%(&! x# 2†9500 “ !
œ 6400 Š4750
4750# 4†4750 ‹
419
œ ˆ 34 ‰ (6400)(4750)
39. Force constant: F œ kx Ê 20 œ k † 1 Ê k œ 20 lb/ft; the work to stretch the spring 1 ft is W œ '0 kx dx œ k'0 x dx œ ’20 x# “ œ 10 ft † lb; the work to stretch the spring an additional foot is 1
1
#
! # x# 20 ’ # “ "
W œ '1 kx dx œ k '1 x dx œ 2
"
2
œ 20 ˆ 4# "# ‰ œ 20 ˆ 3# ‰ œ 30 ft † lb
40. Force constant: F œ kx Ê 200 œ k(0.8) Ê k œ 250 N/m; the 300 N force stretches the spring x œ œ
300 250
F k
œ 1.2 m; the work required to stretch the spring that far is then W œ '0 F(x) dx œ '0 250x dx 1Þ2
1Þ2
œ [125x# ]!"Þ# œ 125(1.2)# œ 180 J 41. We imagine the water divided into thin slabs by planes perpendicular to the y-axis at the points of a partition of the interval [0ß 8]. The typical slab between the planes at y and y ?y has a volume of about ?V œ 1(radius)# (thickness) # œ 1 ˆ 54 y‰ ?y œ 25161 y# ?y ft$ . The force F(y) required to lift this slab is equal to its weight: F(y) œ 62.4 ?V œ
(62.4)(25) 16
1y# ?y lb. The distance through which F(y)
must act to lift this slab to the level 6 ft above the top is about (6 8 y) ft, so the work done lifting the slab is about ?W œ
(62.4)(25) 16
1y# (14 y) ?y ft † lb. The work done
lifting all the slabs from y œ 0 to y œ 8 to the level 6 ft above the top is approximately 8
W¸! !
(62.4)(25) 16
1y# (14 y) ?y ft † lb so the work to pump the water is the limit of these Riemann sums as the norm of
the partition goes to zero: W œ '0
8
œ
(62.4) ˆ 25161 ‰ Š 14 3
$
†8
8% 4‹
(62.4)(25) (16)
1y# (14 y) dy œ
(62.4)(25)1 16
'08 a14y# y$ b dy œ (62.4) ˆ 25161 ‰ ’ 143 y$ y4 “ ) %
!
¸ 418,208.81 ft † lb
42. The same as in Exercise 41, but change the distance through which F(y) must act to (8 y) rather than (6 8 y). Also change the upper limit of integration from 8 to 5. The integral is: W œ '0
5
(62.4)(25)1 16
y# (8 y) dy œ (62.4) ˆ 25161 ‰'0 a8y# y$ b dy œ (62.4) ˆ 25161 ‰ ’ 83 y$ 5
œ (62.4) ˆ 25161 ‰ Š 38 † 5$
5% 4‹
& y% 4 “!
¸ 54,241.56 ft † lb
43. The tank's cross section looks like the figure in Exercise 41 with right edge given by x œ #
horizontal slab has volume ?V œ 1(radius)# (thickness) œ 1 ˆ #y ‰ ?y œ slab is its weight: F(y) œ 60 †
1 4
1 4
10
22,500 ft†lb 275 ft†lb/sec
y œ y# . A typical
y# ?y. The force required to lift this
y# ?y. The distance through which F(y) must act is (2 10 y) ft, so the
work to pump the liquid is W œ 60'0 1(12 y) Š y4 ‹ dy œ 151 ’ 12y 3 to empty the tank is
5 10
#
$
"! y% 4 “!
œ 22,5001 ft † lb; the time needed
¸ 257 sec
44. A typical horizontal slab has volume about ?V œ (20)(2x)?y œ (20) ˆ2È16 y# ‰ ?y and the force required to lift this slab is its weight F(y) œ (57)(20) ˆ2È16 y# ‰ ?y. The distance through which F(y) must act is (6 4 y) ft, so the work to pump the olive oil from the half-full tank is
W œ 57'c4 (10 y)(20) ˆ2È16 y# ‰ dy œ 2880 'c4 10È16 y# dy 1140'c4 a16 y# b 0
0
0
"Î#
(2y) dy
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
420
Chapter 6 Applications of Definite Integrals œ 22,800 † (area of a quarter circle having radius 4) 23 (1140) ’a16 y# b
$Î# !
“
œ 335,153.25 ft † lb
%
œ (22,800)(41) 48,640
strip 45. F œ 'a W † Š depth ‹ † L(y) dy Ê F œ 2 '0 (62.4)(2 y)(2y) dy œ 249.6'0 a2y y# b dy œ 249.6 ’y# b
2
2
# y$ 3 “!
œ (249.6) ˆ4 83 ‰ œ (249.6) ˆ 43 ‰ œ 332.8 lb strip 46. F œ 'a W † Š depth ‹ † L(y) dy Ê F œ '0 75 ˆ 56 y‰ (2y 4) dy œ 75'0 ˆ 53 y 5Î6
b
5Î6
10 3
2y# 4y‰ dy
7 7 # 2 $ ‘ &Î' 50 ‰ 25 ‰ 125 ‰‘ #‰ ˆ 18 œ 75 '0 ˆ 10 dy œ 75 10 ˆ 67 ‰ ˆ 36 ˆ 32 ‰ ˆ 216 3 3 y 2y 3 y 6 y 3 y ! œ (75) 5Î6
œ (75) ˆ 25 9
175 216
250 ‰ 3†#16
‰ œ ˆ 9†75 #16 (25 † 216 175 † 9 250 † 3) œ
strip 47. F œ 'a W † Š depth ‹ † L(y) dy Ê F œ 62.4'0 (9 y) Š2 † b
4
%
œ 62.4 6y$Î# 25 y&Î# ‘ ! œ (62.4) ˆ6 † 8
2 5
Èy 2 ‹
(75)(3075) 9†#16
¸ 118.63 lb.
dy œ 62.4'0 ˆ9y"Î# 3y$Î# ‰ dy 4
‰ (48 † 5 64) œ † 32‰ œ ˆ 62.4 5
(62.4)(176) 5
œ 2196.48 lb
strip 48. Place the origin at the bottom of the tank. Then F œ '0 W † Š depth ‹ † L(y) dy, h œ the height of the mercury column, h
strip depth œ h y, L(y) œ 1 Ê F œ '0 849(h y) " dy œ (849)'0 (h y) dy œ 849’hy h
œ
849 # 2 h .
Now solve
849 # 2 h
h
h
y# # “!
œ 849 Šh#
h# #‹
œ 40000 to get h ¸ 9.707 ft. The volume of the mercury is s2 h œ 12 † 9.707 œ 9.707 ft$ Þ
49. F œ w" '0 (8 y)(2)(6 y) dy w# 'c6 (8 y)(2)(y 6) dy œ 2w" '0 a48 14y y# b dy 2w# '6 a48 2y y# b dy 6
0
œ 2w" ’48y 7y#
' y$ 3 “!
6
2w# ’48y y#
! y$ 3 “ '
0
œ 216w" 360w#
50. (a) F œ 62.4'0 (10 y) ˆ8 y6 ‰ ˆ y6 ‰‘ dy 6
œ
6
62.4 3
' a240 34y y# b dy 0
œ
62.4 3
’240y 17y#
œ 18,720 lb.
' y$ 3 “!
œ
62.4 3
(1440 612 72)
(b) The centroid ˆ 72 ß 3‰ of the parallelogram is located at the intersection of y œ
6 7
x and y œ 65 x
36 5 .
The centroid of
the triangle is located at (7ß 2). Therefore, F œ (62.4)(7)(36) (62.4)(8)(6) œ (300)(62.4) œ 18,720 lb CHAPTER 6 ADDITIONAL AND ADVANCED EXERCISES 1. V œ 1 'a cf(x)d# dx œ b# ab Ê 1'a cf(t)d# dt œ x# ax for all x a Ê 1 [f(x)]# œ 2x a Ê f(x) œ „ É 2x1 a b
x
2. V œ 1 '0 [f(x)]# dx œ a# a Ê 1 '0 [f(t)]# dt œ x# x for all x a Ê 1[f(x)]# œ 2x 1 Ê f(x) œ „ É 2x1 1 a
x
3. s(x) œ Cx Ê '0 È1 [f w (t)]# dt œ Cx Ê È1 [f w (x)]# œ C Ê f w (x) œ ÈC# 1 for C 1 x
Ê f(x) œ '0 ÈC# 1 dt k. Then f(0) œ a Ê a œ 0 k Ê f(x) œ '0 ÈC# 1 dt a Ê f(x) œ xÈC# 1 a, x
x
where C 1.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 6 Additional and Advanced Exercises
421
4. (a) The graph of f(x) œ sin x traces out a path from (!ß !) to (!ß sin !) whose length is L œ '0 È1 cos# ) d). !
The line segment from (0ß 0) to (!ß sin !) has length È(! 0)# (sin ! 0)# œ È!# sin# !. Since the shortest distance between two points is the length of the straight line segment joining them, we have ! immediately that ' È1 cos# ) d) È!# sin# ! if 0 ! Ÿ 1 . #
0
(b) In general, if y œ f(x) is continuously differentiable and f(0) œ 0, then '0 È1 [f w (t)]# dt È!# f # (!) !
for ! 0. 5. From the symmetry of y œ 1 xn , n even, about the y-axis for 1 Ÿ x Ÿ 1, we have x œ 0. To find y œ MMx , we n use the vertical strips technique. The typical strip has center of mass: (µ x ßµ y ) œ ˆxß 1 2 x ‰ , length: 1 xn , width: dx, area: dA œ a1 xn b dx, mass: dm œ 1 † dA œ a1 xn b dx. The moment of the strip about the 1 1 n # n # " nb1 2n b 1 x-axis is µ y dm œ a1 x b dx Ê M œ ' a1 x b dx œ 2' " a1 2xn x2n b dx œ x 2x x ‘ #
œ1
2 n1
" #n 1
œ
x c1 # (n 1)(2n 1) 2(2n ") (n 1) (n 1)(#n 1)
œ
0 # 2n# 3n 1 4n 2 n 1 (n 1)(#n 1)
Also, M œ 'c1 dA œ 'c1 a1 xn b dx œ 2 '0 a1 xn b dx œ 2 x 1
yœ
Mx M
œ
1
#
2n (n 1)(2n 1)
†
1
(n 1) 2n
œ
n 2n 1
xn b 1 ‘ " n1 !
n1
œ
2n# (n 1)(#n 1)
œ 2 ˆ1
" ‰ n1
#n 1 !
. œ
2n n1.
Therefore,
Ê ˆ!ß #n n 1 ‰ is the location of the centroid. As n Ä _, y Ä
" #
so
the limiting position of the centroid is ˆ!ß "# ‰ . 6. Align the telephone pole along the x-axis as shown in the accompanying figure. The slope of the top length of pole is 9 ‰ ˆ 14.5 " 81 81 œ 8"1 † 40 † (14.5 9) œ 815.5 †40 40 11 ‰ y œ 891 8111†80 x œ 8"1 ˆ9 80 x is an
œ
11 81†80 .
Thus,
equation of the
line representing the top of the pole. Then, My œ 'a x † 1y# dx œ 1 '0 x 8"1 ˆ9 b
40
11 80
# x‰‘ dx
b 11 ‰# '040 x ˆ9 80 x dx; M œ 'a 1y# dx 40 40 ‰‘# dx œ 64"1 ' ˆ9 11 ‰# dx. œ 1 '0 8"1 ˆ9 11 80 x 80 x 0
œ
" 641
My M
Thus, x œ
¸
129,700 5623.3
¸ 23.06 (using a calculator to compute
the integrals). By symmetry about the x-axis, y œ 0 so the center of mass is about 23 ft from the top of the pole. 7. (a) Consider a single vertical strip with center of mass (µ x ßµ y ). If the plate lies to the right of the line, then µ µ b) $ dA Ê the plate's first moment the moment of this strip about the line x œ b is (x b) dm œ (x about x œ b is the integral ' (x b)$ dA œ ' $ x dA ' $ b dA œ My b$ A. (b) If the plate lies to the left of the line, the moment of a vertical strip about the line x œ b is ab µ x b dm œ ab µ x b $ dA Ê the plate's first moment about x œ b is ' (b x)$ dA œ ' b$ dA ' $ x dA œ b$ A My . 8. (a) By symmetry of the plate about the x-axis, y œ 0. A typical vertical strip has center of mass: (µ x ßµ y ) œ (xß 0), length: 4Èax, width: dx, area: 4Èax dx, mass: dm œ $ dA œ kx † 4Èax dx, for some a proportionality constant k. The moment of the strip about the y-axis is M œ ' µ x dm œ ' 4kx# Èax dx y
œ 4kÈa'0 x&Î# dx œ 4kÈa 27 x(Î# ‘ 0 œ 4ka"Î# † 27 a(Î# œ a
a
œ 4kÈa'0 x$Î# dx œ 4kÈa 25 x&Î# ‘ 0 œ 4ka"Î# † 25 a&Î# œ a
a
8ka 7
%
8ka$ 5
0
. Also, M œ ' dm œ '0 4kxÈax dx a
. Thus, x œ
My M
œ
8ka% 7
†
5 8ka$
œ
5 7
a
‰ Ê (xß y) œ ˆ 5a 7 ß 0 is the center of mass. y#
# # a (b) A typical horizontal strip has center of mass: (µ x ßµ y ) œ Œ 4a # ß y œ Š y 8a4a ß y‹ , length: a
width: dy, area: Ša
y# 4a ‹
dy, mass: dm œ $ dA œ kyk Ša
y# 4a ‹
dy. Thus, Mx œ ' µ y dm
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
y# 4a
,
422
Chapter 6 Applications of Definite Integrals œ 'c2a y kyk Ša 2a
œ 'c2a Šay# 0
%
32a& 20a
œ 8a3
y# 4a ‹
0
dy '0 Šay# 2a
y% 4a ‹ 8a% 3
dy œ 'c2a y# Ša
#
œ
%
&
'c2a a16a y y b dy
œ
" 32a#
’8a% † 4a#
64a' 6 “
" 32a#
'0
2a
" 32a#
2a
#
c2a
" 3 #a #
a16a% y y& b dy œ 64a' 6 “
" 16a#
œ
’8a% y# 32a' 3 ‹
Š32a'
œ
œ2†
16a% 4 ‹
#
Š2a † 4a
" #a
œ
! y' 6 “ #a
1 3#a#
’8a% y#
† 32 a32a' b œ
4 3
#a y' 6 “!
a% ;
'c2a2a kyk a4a# y# b dy
" 4a
#a
#
dy
" 16a#
%
" 4a
#a y& “ #0a !
y# 4a ‹
kyk Ša
'c02a a4a# y y$ b dy 4a" '02a a4a# y y$ b dy œ 4a" ’2a# y# y4 “ !
" 4a
yœ
#
4a# 8a ‹
’ 3a y$
' kyk a16a% y% b dy
#
’8a% † 4a#
M œ ' dm œ 'c2a kyk Š 4a 4ay ‹ dy œ
y& #0a “ #a
dy
2a
#
" 3 #a #
#
y# 4a ‹
!
dy œ ’ 3a y$
'c2a2a kyk ay# 4a# b Š 4a 4a y ‹ dy œ 32a"
2a
2a
" 8a
0
dy '0 y# Ša
œ 0; My œ ' µ x dm œ 'c2a Š y
32a& #0a
œ
œ
y% 4a ‹
y# 4a ‹
%
%
$
a8a 4a b œ 2a . Therefore, x œ
" 4a
’2a# y#
œ ˆ 34 a% ‰ ˆ 2a"$ ‰ œ
My M
#a y% 4 “!
2a 3
and
œ 0 is the center of mass.
Mx M
9. (a) On [0ß a] a typical @/3-+6 strip has center of mass: (µ x ßµ y ) œ Šx,
È b # x # È a# x# ‹, #
length: Èb# x# Èa# x# , width: dx, area: dA œ ŠÈb# x# Èa# x# ‹ dx, mass: dm œ $ dA œ $ ŠÈb# x# Èa# x# ‹ dx. On [aß b] a typical @/3-+6 strip has center of mass: È # # (µ x ßµ y ) œ Šxß b # x ‹ , length: Èb# x# , width: dx, area: dA œ Èb# x# dx,
mass: dm œ $ dA œ $ Èb# x# dx. Thus, Mx œ ' µ y dm œ '0
a
" #
ŠÈb# x# Èa# x# ‹ $ ŠÈb# x# Èa# x# ‹ dx 'a
b
" #
Èb# x# $ Èb# x# dx
œ
$ #
'0a cab# x# b aa# x# bd dx #$ 'ab ab# x# b dx œ #$ '0a ab# a# b dx #$ 'ab ab# x# b dx
œ
$ #
cab# a# b xd ! #$ ’b# x
œ
$ #
aab# a$ b #$ Š 23 b$ ab#
a
b
x$ 3 “a
œ a$ 3‹
$ #
b$ 3‹
cab# a# b ad #$ ’Šb$
œ
$ b$ 3
$ a$ 3
œ $ Šb
$
a$ 3 ‹;
a$ 3 ‹“
Š b# a
My œ ' µ x dm
œ '0 x$ ŠÈb# x# Èa# x# ‹ dx 'a x$ Èb# x# dx a
b
œ $ '0 x ab# x# b a
œ
$ #
”
2 ab # x # b 3
$Î#
dx $ '0 x aa# x# b a
"Î# a
$ 2 aa • #”
#
$Î#
x# b 3
# $Î#
#
œ ’ab a b
# $Î#
ab b
a
dx $ 'a x ab# x# b
$ 2 ab • #”
b
#
!
0
$ 3
"Î#
# $Î#
$ 3
“ ’0 aa b
• a
$ 3
“ ’0 ab# a# b #
#
$Î#
We calculate the mass geometrically: M œ $ A œ $ Š 14b ‹ $ Š 14a ‹ œ œ
$ ab $ a $ b 3
yœ (b) lim
œ
Mx M 4
b Ä a 31
†
4 $1 ab# a# b
4 aa# abb# b 31(ab)
Ša
#
ab b# ‹ ab
œ
4 31
$
$
a Š bb# a# ‹ œ
dx
b
$Î#
x# b 3
"Î#
4 (b a) aa# ab b# b 31 (b a)(b a)
“œ
$1 4
$ b$ 3
$ a$ 3
œ
$ ab $ a $ b 3
ab# a# b . Thus, x œ
œ Mx ; My M
œ
4 aa# ab b# b 31(a b)
2a 1
2a ‰ Ê (xß y) œ ˆ 2a 1 ß 1 is the limiting
; likewise
. œ ˆ 341 ‰ Š a
#
a# a# ‹ aa
#
œ ˆ 341 ‰ Š 3a 2a ‹ œ
position of the centroid as b Ä a. This is the centroid of a circle of radius a (and we note the two circles coincide when b œ a).
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 6 Additional and Advanced Exercises 10. Since the area of the traingle is 36, the diagram may be labeled as shown at the right. The centroid of the triangle is ˆ 3a , 24 ‰ a . The shaded portion is 144 36 œ 108. Write ax, yb for the centroid of the remaining region. The centroid of the whole square is obviously a6, 6b. Think of the square as a sheet of uniform density, so that the centroid of the square is the average of the centroids of the two regions, weighted by area: 'œ
$'ˆ 3a ‰ "!)axb "%%
and ' œ
‰ $'ˆ 24 a "!)ayb "%%
which we solve to get x œ )
a *
and y œ
)a a " b . a
Set
x œ 7 in. (Given). It follows that a œ *, whence y œ œ
7 "*
'% *
in. The distances of the centroid ax, yb from the other sides are easily computed. (Note that if we set y œ 7 in.
above, we will find x œ 7 "* .) 11. y œ 2Èx Ê ds œ É "x 1 dx Ê A œ '0 2Èx É "x 1 dx œ 3
4 3
(1 x)$Î# ‘ $ œ !
28 3
12. This surface is a triangle having a base of 21a and a height of 21ak. Therefore the surface area is " # # # (21a)(21ak) œ 21 a k. d# x dt#
13. F œ ma œ t# Ê
œaœ
t# m
Ê vœ
x œ 0 when t œ 0 Ê C" œ 0 Ê x œ W œ ' F dx œ '0
Ð12mhÑ"Î%
œ
(12mh)$Î# 18m
œ
F(t) †
12mh†È12mh 18m
œ
2h 3
dx dt
dx t$ dt œ 3m C; v œ 0 when t œ 0 Ê t% "Î% . 12m . Then x œ h Ê t œ (12mh)
dt œ '0
Ð12mhÑ"Î%
† 2È3mh œ
14. Converting to pounds and feet, 2 lb/in œ
t# †
t$ 3m
dt œ
4h 3
È3mh
†
12 in 1 ft
2 lb 1 in
" 3m
'
’ t6 “
Ð12mh)"Î%
0
Cœ0 Ê
dx dt
œ
t$ 3m
Ê xœ
The work done is
" ‰ œ ˆ 18m (12mh)'Î%
œ 24 lb/ft. Thus, F œ 24x Ê W œ '0
1Î2
24x dx
"Î# " " ‰ œ c12x# d ! œ 3 ft † lb. Since W œ "# mv!# "# mv"# , where W œ 3 ft † lb, m œ ˆ 10 lb‰ ˆ 3# ft/sec # " œ 320 slugs, and v" œ 0 ft/sec, we have 3 œ ˆ #" ‰ ˆ 3#"0 v#! ‰ Ê v!# œ 3 † 640. For the projectile height, s œ 16t# v! t (since s œ 0 at t œ 0) Ê ds dt œ v œ 32t v! . At the top of the ball's path, v œ 0 Ê #
and the height is s œ 16 ˆ 3v#! ‰ v! ˆ 3v#! ‰ œ
v!# 64
œ
3†640 64
œ 30 ft.
15. The submerged triangular plate is depicted in the figure at the right. The hypotenuse of the triangle has slope 1 Ê y (2) œ (x 0) Ê x œ (y 2) is an equation of the hypotenuse. Using a typical horizontal strip, the fluid strip strip pressure is F œ ' (62.4) † Š depth ‹ † Š length ‹ dy
c2
c2
œ 'c6 (62.4)(y)[(y 2)] dy œ 62.4 'c6 ay# 2yb dy $
œ 62.4 ’ y3 y# “
# '
‰‘ œ (62.4) ˆ 83 4‰ ˆ 216 3 36
‰ œ (62.4) ˆ 208 3 32 œ
(62.4)(112) 3
t% 12m
¸ 2329.6 lb
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
tœ
v! 3#
C" ;
423
424
Chapter 6 Applications of Definite Integrals
16. Consider a rectangular plate of length j and width w. The length is parallel with the surface of the fluid of weight density =. The force on one side of the plate is F œ ='cw (y)(j) dy œ =j ’ y# “ 0
#
!
œ
w
=jw# #
. The = w
average force on one side of the plate is Fav œ œ
= w
#
’ y# “
!
=w #
œ
w
. Therefore the force
'c0w (y)dy
=jw# #
‰ œ ˆ =w # (jw) œ (the average pressure up and down) † (the area of the plate). 17. (a) We establish a coordinate system as shown. A typical horizontal strip has: center of pressure: (µ x ßµ y ) b ˆ ‰ œ # ß y , length: L(y) œ b, width: dy, area: dA
œ b dy, pressure: dp œ = kyk dA œ =b kyk dy 0 0 Ê Fx œ ' µ y dp œ 'ch y † =b kyk dy œ =b 'ch y# dy $
œ =b ’ y3 “
!
œ =b ’0 Š 3h ‹“ œ
=bh$ 3
'c0h = kyk L(y) dy œ =b 'c0h
F œ ' dp œ œ
$
h
# ! =b ’ y# “ h
œ =b ’0
h# #“
=bh# #
œ
;
y dy $
. Thus, y œ
œ
Fx F
Š =3bh ‹
#
Š =bh # ‹
œ
2h 3
Ê the distance below the surface is
(b) A typical horizontal strip has length L(y). By similar triangles from the figure at the right,
L(y) b
œ
y a h
Ê L(y) œ bh (y a). Thus, a typical strip has center of pressure: (µ x ßµ y ) œ (µ x ß y), length: L(y) œ bh (y a), width: dy, area: dA œ bh (y a) dy, pressure: dp œ = kyk dA œ =(y) ˆ bh ‰ (y a) dy œ =b ay# ayb dy Ê F œ ' µ y dp h
x
a
œ 'cÐahÑ y † %
=b h
#
ay ayb dy œ
'ÐaahÑ
=b h
a ay$ 3 “ cÐahÑ
ay$ ay# b dy
œ
=b h
’ y4
œ
=b h
’Š a4
œ œ
=b 12h =b 12h
œ
=bh 12
œ
=b h
’Š 3a
œ
=b h
’a
œ
=b 6h
a6a# h 6ah# 2h$ 6a# h 3ah# b œ
œ
%
a% 3‹
%
Š (a 4 h)
a(a h)$ ‹“ 3
œ
=b h
’a
%
(a h)% 4
a% a(a h)$ “ 3
c3 aa% aa% 4a$ h 6a# h# 4ah$ h% bb 4 aa% a aa$ 3a# h 3ah# h$ bbd a12a$ h 12a# h# 4ah$ 12a$ h 18a# h# 12ah$ 3h% b œ a6a# 8ah 3h# b ; F œ ' dp œ ' = kyk L(y) dy œ $
$
a$ #‹
$
Š (a 3 h)
3a# h 3ah# h$ a$ 3
ˆ 1=#bh ‰ a6a# 8ah 3h# b ˆ =6bh ‰ (3a 2h)
6a# 8ah 3h# 6a 4h
a(a h)# ‹“ #
œ
=b h
a$ aa$ 2a# h ah# b “ #
‰ 6a œ ˆ " # Š
#
=b 6h
8ah 3h# ‹ 3a 2h
$
’ (a h)3
œ
=b 6h
a$
a
=b 12h
a6a# h# 8ah$ 3h% b
=b h
'cÐahÑ
a$ a(a h)# “ 2
ay# ayb dy œ
=b h
$
’ y3
c2 a3a# h 3ah# h$ b 3 a2a# h ah# bd
a3ah# 2h$ b œ
=bh 6
(3a 2h). Thus, y œ
Ê the distance below the surface is
.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Fx F
a ay# 2 “ ÐahÑ
2 3
h.
CHAPTER 7 TRANSCENDENTAL FUNCTIONS 7.1 INVERSE FUNCTIONS AND THEIR DERIVATIVES 1. Yes one-to-one, the graph passes the horizontal test. 2. Not one-to-one, the graph fails the horizontal test. 3. Not one-to-one since (for example) the horizontal line y œ # intersects the graph twice. 4. Not one-to-one, the graph fails the horizontal test. 5. Yes one-to-one, the graph passes the horizontal test 6. Yes one-to-one, the graph passes the horizontal test 7. Domain: 0 x Ÿ 1, Range: 0 Ÿ y
9. Domain: 1 Ÿ x Ÿ 1, Range: 1# Ÿ y Ÿ
8. Domain: x 1, Range: y 0
1 #
10. Domain: _ x _, Range: 1# y Ÿ
11. The graph is symmetric about y œ x.
(b) y œ È1 x# Ê y# œ 1 x# Ê x# œ 1 y# Ê x œ È1 y# Ê y œ È1 x# œ f " (x)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
1 #
426
Chapter 7 Transcendental Functions
12. The graph is symmetric about y œ x.
yœ
" x
Ê xœ
" y
Ê yœ
" x
œ f " (x)
13. Step 1: y œ x# 1 Ê x# œ y 1 Ê x œ Èy 1 Step 2: y œ Èx 1 œ f " (x) 14. Step 1: y œ x# Ê x œ Èy, since x Ÿ !. Step 2: y œ Èx œ f " (x) 15. Step 1: y œ x$ 1 Ê x$ œ y 1 Ê x œ (y 1)"Î$ Step 2: y œ $Èx 1 œ f " (x) 16. Step 1: y œ x# 2x 1 Ê y œ (x 1)# Ê Èy œ x 1, since x 1 Ê x œ 1 Èy Step 2: y œ 1 Èx œ f " (x) 17. Step 1: y œ (x 1)# Ê Èy œ x 1, since x 1 Ê x œ Èy 1 Step 2: y œ Èx 1 œ f " (x) 18. Step 1: y œ x#Î$ Ê x œ y$Î# Step 2: y œ x$Î# œ f " (x) 19. Step 1: y œ x& Ê x œ y"Î& Step 2: y œ &Èx œ f " (x); Domain and Range of f " : all reals; &
f af " (x)b œ ˆx"Î& ‰ œ x and f " (f(x)) œ ax& b
"Î&
œx
"Î%
œx
20. Step 1: y œ x% Ê x œ y"Î% Step 2: y œ %Èx œ f " (x); Domain of f " : x 0, Range of f " : y 0; %
f af " (x)b œ ˆx"Î% ‰ œ x and f " (f(x)) œ ax% b
21. Step 1: y œ x$ 1 Ê x$ œ y 1 Ê x œ (y 1)"Î$ Step 2: y œ $Èx 1 œ f " (x); Domain and Range of f " : all reals; $
f af " (x)b œ ˆ(x 1)"Î$ ‰ 1 œ (x 1) 1 œ x and f " (f(x)) œ aax$ 1b 1b
"Î$
œ ax$ b
"Î$
œx
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 7.1 Inverse Functions and Their Derivatives 22. Step 1: y œ
" #
x
" #
Ê
7 #
"
xœy
7 #
Ê x œ 2y 7
Step 2: y œ 2x 7 œ f (x); Domain and Range of f " : all reals; f af " (x)b œ "# (2x 7) 7# œ ˆx 7# ‰ 23. Step 1: y œ Step 2: y œ
" x#
Ê x# œ
" y
" Èx
œ f " (x)
Ê xœ
7 #
œ x and f " (f(x)) œ 2 ˆ "# x 7# ‰ 7 œ (x 7) 7 œ x
" Èy
Domain of f " : x 0, Range of f " : y 0; f af " (x)b œ "" # œ "" œ x and f " (f(x)) œ Š Èx ‹
24. Step 1: y œ
" x$
Ê x$ œ
" x"Î$ "
Step 2: y œ Domain of f f af " (x)b œ
Šx‹
" y
Ê xœ
(c)
26. (a) y œ
" 5
" $ ax"Î$ b
" x"
œ
œ 2,
df " dx ¹ xœ1
x7 Ê
df ¸ dx xœ1
(c)
œ x since x 0
" y"Î$
œ x and f " (f(x)) œ ˆ x"$ ‰
" 5
œ
" df "
œ 5,
dx
¹
œ 4,
df " dx ¹ xœ3
œ ˆ x" ‰
"
œx
(b) x #
3 #
xœy7
xœ$%Î&
"Î$
" #
"
(b)
(x) œ 5x 35
œ5
27. (a) y œ 5 4x Ê 4x œ 5 y Ê x œ 54 y4 Ê f " (x) œ df ¸ dx xœ1Î#
" Š "x ‹
: x Á 0, Range of f " : y Á 0;
Ê x œ 5y 35 Ê f (c)
œ
œ $É x" œ f " (x);
25. (a) y œ 2x 3 Ê 2x œ y 3 Ê x œ y# 3# Ê f " (x) œ df ¸ dx xœ1
" É x"#
œ
(b) 5 4
x 4
" 4
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
427
428
Chapter 7 Transcendental Functions " #
28. (a) y œ 2x# Ê x# œ Ê xœ (c)
df ¸ dx xœ&
" È2
(b)
y
Èy Ê f
"
(x) œ
È x#
œ 4xk xœ5 œ 20,
df " dx ¹ xœ&0
œ
" #È 2
x"Î# ¹
xœ50
" #0
œ
$ $ 29. (a) f(g(x)) œ ˆ $Èx‰ œ x, g(f(x)) œ Èx$ œ x
w
#
w
(b)
w
(c) f (x) œ 3x Ê f (1) œ 3, f (1) œ 3; gw (x) œ 3" x#Î$ Ê gw (1) œ 3" , gw (1) œ
" 3
(d) The line y œ 0 is tangent to f(x) œ x$ at (!ß !); the line x œ 0 is tangent to g(x) œ $Èx at (0ß 0)
30. (a) h(k(x)) œ
" 4
ˆ(4x)"Î$ ‰$ œ x,
k(h(x)) œ Š4 † (c) hw (x) œ w
k (x) œ
x$ 4‹
"Î$
(b)
œx
#
3x w w 4 Ê h (2) œ 3, h (2) 4 #Î$ Ê kw (2) œ "3 , 3 (4x)
œ 3; kw (2) œ
(d) The line y œ 0 is tangent to h(x) œ
x$ 4
" 3
at (!ß !);
the line x œ 0 is tangent to k(x) œ (4x)"Î$ at (!ß !)
œ 3x# 6x Ê
31.
df dx
33.
df " dx ¹ x œ 4
df " dx ¹ x œ f(3)
df " dx ¹ x œ f(2)
œ
35. (a) y œ mx Ê x œ
" m
œ
(b) The graph of y œ f 36. y œ mx b Ê x œ
y m
"
df dx
º
œ
xœ2
"
df dx
œ
º
xœ3
" ˆ 3" ‰
œ3
y Ê f " (x) œ "
" 9
œ
" m
œ 2x 4 Ê
32.
df dx
34.
dg" dx ¹x œ 0
b m
dg" dx ¹ x œ f(0)
œ
"
dg dx
º
œ
xœ0
"
df dx
º
œ
xœ5
œ
" 6
" 2
x
(x) is a line through the origin with slope
œ
df " dx ¹ x œ f(5)
Ê f " (x) œ
" m
x
b m;
" m.
the graph of f " (x) is a line with slope
" m
and y-intercept mb .
37. (a) y œ x 1 Ê x œ y 1 Ê f " (x) œ x 1 (b) y œ x b Ê x œ y b Ê f " (x) œ x b (c) Their graphs will be parallel to one another and lie on opposite sides of the line y œ x equidistant from that line.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 7.1 Inverse Functions and Their Derivatives 38. (a) y œ x 1 Ê x œ y 1 Ê f " (x) œ 1 x; the lines intersect at a right angle (b) y œ x b Ê x œ y b Ê f " (x) œ b x; the lines intersect at a right angle (c) Such a function is its own inverse.
39. Let x" Á x# be two numbers in the domain of an increasing function f. Then, either x" x# or x" x# which implies f(x" ) f(x# ) or f(x" ) f(x# ), since f(x) is increasing. In either case, f(x" ) Á f(x# ) and f is one-to-one. Similar arguments hold if f is decreasing. 40. f(x) is increasing since x# x" Ê
" 3
x#
5 6
" 3
x" 56 ;
df dx
œ
" 3
41. f(x) is increasing since x# x" Ê 27x$# 27x"$ ; y œ 27x$ Ê x œ df dx
œ 81x# Ê
df " dx
œ
" ¸ 81x# 13 x"Î$
œ
" 9x#Î$
œ
" 9
df " dx
Ê " 3
œ
df dx
œ 24x# Ê
dx
œ
" ¸ 24x# 12 Ð1xÑ"Î$
œ
œ3
y"Î$ Ê f " (x) œ
" 3
x"Î$ ;
x#Î$
42. f(x) is decreasing since x# x" Ê 1 8x$# 1 8x"$ ; y œ 1 8x$ Ê x œ df "
" ˆ "3 ‰
" 6(" x)#Î$
" #
(1 y)"Î$ Ê f " (x) œ
" #
(1 x)"Î$ ;
œ "6 (1 x)#Î$
43. f(x) is decreasing since x# x" Ê (1 x# )$ (1 x" )$ ; y œ (1 x)$ Ê x œ 1 y"Î$ Ê f " (x) œ 1 x"Î$ ; df dx
œ 3(1 x)# Ê
df " dx
œ
" 3(1 x)# ¹ 1cx"Î$ &Î$
44. f(x) is increasing since x# x" Ê x# df dx
œ
5 3
x#Î$ Ê
df " dx
œ
5 3
" ¹ x#Î$ x$Î&
œ
3 5x#Î&
œ
" 3x#Î$
œ "3 x#Î$
&Î$
x" ; y œ x&Î$ Ê x œ y$Î& Ê f " (x) œ x$Î& ; œ
3 5
x#Î&
45. The function g(x) is also one-to-one. The reasoning: f(x) is one-to-one means that if x" Á x# then f(x" ) Á f(x# ), so f(x" ) Á f(x# ) and therefore g(x" ) Á g(x# ). Therefore g(x) is one-to-one as well. 46. The function h(x) is also one-to-one. The reasoning: f(x) is one-to-one means that if x" Á x# then f(x" ) Á f(x# ), so f(x"" ) Á f(x"# ) , and therefore h(x" ) Á h(x# ). 47. The composite is one-to-one also. The reasoning: If x" Á x# then g(x" ) Á g(x# ) because g is one-to-one. Since g(x" ) Á g(x# ), we also have f(g(x" )) Á f(g(x# )) because f is one-to-one. Thus, f ‰ g is one-to-one because x" Á x# Ê f(g(x" )) Á f(g(x# )). 48. Yes, g must be one-to-one. If g were not one-to-one, there would exist numbers x" Á x# in the domain of g with g(x" ) œ g(x# ). For these numbers we would also have f(g(x" )) œ f(g(x# )), contradicting the assumption that f ‰ g is one-to-one.
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429
430
Chapter 7 Transcendental Functions
49. The first integral is the area between f(x) and the x-axis over a Ÿ x Ÿ b. The second integral is the area between f(x) and the y-axis for f(a) Ÿ y Ÿ f(b). The sum of the integrals is the area of the larger rectangle with corners at (0ß 0), (bß 0), (bß f(b)) and (0ß f(b)) minus the area of the smaller rectangle with vertices at (0ß 0), (aß 0), (aß f(a)) and (0ß f(a)). That is, the sum of the integrals is bf(b) af(a).
50. f w axb œ
acx dba aax bbc acx db#
œ
ad bc . acx db#
Thus if ad bc Á !, f w axb is either always positive or always negative. Hence faxb is
either always increasing or always decreasing. If follows that faxb is one-to-one if ad bc Á !. 51. (g ‰ f)(x) œ x Ê g(f(x)) œ x Ê gw (f(x))f w (x) œ 1 52. W(a) œ 'f(a) 1 ’af " (y)b a# “ dy œ 0 œ 'a 21x[f(a) f(x)] dx œ S(a); Ww (t) œ 1’af " (f(t))b a# “ f w (t) f(a)
a
#
#
œ 1 at# a# b f w (t); also S(t) œ 21f(t)'a x dx 21'a xf(x) dx œ c1f(t)t# 1f(t)a# d 21'a xf(x) dx t
t
t
Ê Sw (t) œ 1t# f w (t) 21tf(t) 1a# f w (t) 21tf(t) œ 1 at# a# b f w (t) Ê Ww (t) œ Sw (t). Therefore, W(t) œ S(t) for all t − [aß b]. 53-60. Example CAS commands: Maple: with( plots );#53 f := x -> sqrt(3*x-2); domain := 2/3 .. 4; x0 := 3; Df := D(f); # (a) plot( [f(x),Df(x)], x=domain, color=[red,blue], linestyle=[1,3], legend=["y=f(x)","y=f '(x)"], title="#53(a) (Section 7.1)" ); q1 := solve( y=f(x), x ); # (b) g := unapply( q1, y ); m1 := Df(x0); # (c) t1 := f(x0)+m1*(x-x0); y=t1; m2 := 1/Df(x0); # (d) t2 := g(f(x0)) + m2*(x-f(x0)); y=t2; domaing := map(f,domain); # (e) p1 := plot( [f(x),x], x=domain, color=[pink,green], linestyle=[1,9], thickness=[3,0] ): p2 := plot( g(x), x=domaing, color=cyan, linestyle=3, thickness=4 ): p3 := plot( t1, x=x0-1..x0+1, color=red, linestyle=4, thickness=0 ): p4 := plot( t2, x=f(x0)-1..f(x0)+1, color=blue, linestyle=7, thickness=1 ): p5 := plot( [ [x0,f(x0)], [f(x0),x0] ], color=green ): display( [p1,p2,p3,p4,p5], scaling=constrained, title="#53(e) (Section 7.1)" ); Mathematica: (assigned function and values for a, b, and x0 may vary) If a function requires the odd root of a negative number, begin by loading the RealOnly package that allows Mathematica to do this. See section 2.5 for details. (1) œ '0 ect dt œ lim
bÄ_
" 3
x$ ln (ax) 9" x$ C
'0b ect dt œ
lim cect d b0 œ lim e"b (1)‘ œ 0 1 œ 1
bÄ_
bÄ_
(b) u œ tx , du œ xtxc1 dt; dv œ ect dt, v œ ect ; x œ fixed positive real _
_
Ê >(x 1) œ '0 tx et dt œ lim ctx et d b0 x '0 tx1 et dt œ lim ˆ beb 0x e! ‰ x>(x) œ x>(x) bÄ_
x
bÄ_
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 8 Additional and Advanced Exercises (c) >(n 1) œ n>(n) œ n!: n œ 0: >(0 1) œ >(1) œ 0!; n œ k: Assume >(k 1) œ k! n œ k 1: >(k 1 1) œ (k 1) >(k 1) œ (k 1)k! œ (k 1)! Thus, >(n 1) œ n>(n) œ n! for every positive integer n.
for some k 0; from part (b) induction hypothesis definition of factorial
x n n 50. (a) >(x) ¸ ˆ xe ‰ É 2x1 and n>(n) œ n! Ê n! ¸ n ˆ ne ‰ É 2n1 œ ˆ ne ‰ È2n1
(b)
n 10 20 30 40 50 60
ˆ ne ‰n È2n1 3598695.619 2.4227868 ‚ 10") 2.6451710 ‚ 10$# 8.1421726 ‚ 10%( 3.0363446 ‚ 10'% 8.3094383 ‚ 10)"
calculator 3628800 2.432902 ‚ 10") 2.652528 ‚ 10$# 8.1591528 ‚ 10%( 3.0414093 ‚ 10'% 8.3209871 ‚ 10)"
(c)
n 10
ˆ ne ‰n È2n1 3598695.619
ˆ ne ‰n È2n1 e1Î12n 3628810.051
calculator 3628800
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583
584
Chapter 8 Techniques of Integration
NOTES:
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
CHAPTER 9 FURTHER APPLICATIONS OF INTEGRATION 9.1 SLOPE FIELDS AND SEPARABLE DIFFERENTIAL EQUATIONS 1. (a) y œ ex Ê y w œ ex Ê 2y w 3y œ 2 aex b 3ex œ ex (b) y œ ex e3xÎ2 Ê y w œ ex 3# e3xÎ2 Ê 2y w 3y œ 2 ˆex 3# e3xÎ2 ‰ 3 aex e3xÎ2 b œ ex (c) y œ ex Ce3xÎ2 Ê y w œ ex 3# Ce3xÎ2 Ê 2y w 3y œ 2 ˆex 3# Ce3xÎ2 ‰ 3 aex Ce3xÎ2 b œ ex 2. (a) y œ "x Ê y w œ
" x#
(b) y œ x " 3 Ê y w œ (c) y œ 3. y œ
" xC
'1x
" x
et t
Ê yw œ
#
œ ˆ x" ‰ œ y# " (x 3)#
" (x C)#
#
œ ’ (x " 3) “ œ y# #
œ x " C ‘ œ y#
dt Ê yw œ x"# '1
x t e
t
dt ˆ x" ‰ ˆ ex ‰ Ê x# y w œ '1
x t e
x
t
dt ex œ x Š x"
'1x et
t
dt‹ ex œ xy ex
Ê x# y w xy œ ex 4. y œ
" È 1 x%
'1x È1 t% dt $
Ê y w œ Š 12xx% ‹ Š È
" 1 x%
Ê y w œ #" –
' x È1 t% dt È "
4x$ $— 1 È Š 1 x% ‹
'1x È1 t% dt‹ 1
1 x%
ŠÈ 1 x% ‹
$
Ê y w œ Š 12xx% ‹ y 1 Ê y w
2x$ 1 x%
†yœ1
5. y œ ecx tan" a2ex b Ê y w œ ecx tan" a2ex b ecx ’ 1 a"2ex b# “ a2ex b œ ecx tan" a2ex b Ê y w œ y
2 1 4e2x
Ê yw y œ
2 1 4e2x
2 1 4e2x
; y( ln 2) œ ecÐ ln 2Ñ tan" a2e ln 2 b œ 2 tan" 1 œ 2 ˆ 14 ‰ œ
1 #
# # # # # 6. y œ (x 2) ex Ê y w œ ex ˆ2xex ‰ (x 2) Ê y w œ ex 2xy; y(2) œ (2 2) e2 œ 0
7. y œ
Ê xy y œ 8. y œ
x ln x
dy dx "Î#
9. 2Èxy œ 2x 10.
dy dx
x sin x cos x Ê y w œ sinx x x# 1/2) sin x; y ˆ 1# ‰ œ cos(1(/2) œ0
Ê yw œ
cos x x w
Ê yw œ
ln x x Š "x ‹ (ln x)#
Ê yw œ
" ln x
"x ˆ cosx x ‰ Ê y w œ sinx x
" (ln x)#
Ê x# y w œ
œ 1 Ê 2x"Î# y"Î# dy œ dx Ê 2y"Î# dy œ x"Î# dx Ê C" Ê
2 3
y$Î# x"Î# œ C, where C œ
" #
"Î#
" 3
x# (ln x)#
Ê xy w œ sin x y
Ê x# y w œ xy y# ; y(e) œ
' 2y"Î# dy œ ' x"Î# dx
e ln e
Ê 2 ˆ 23 y$Î# ‰
C"
œ x# Èy Ê dy œ x# y"Î# dx Ê y"Î# dy œ x# dx Ê
Ê 2y
x# ln x
y x
' y"Î# dy œ ' x# dx
Ê 2y"Î# œ
x$ 3
C
$
x œC
' ey dy œ ' ex dx
11.
dy dx
œ excy Ê dy œ ex ecy dx Ê ey dy œ ex dx Ê
12.
dy dx
œ 3x# ey Ê dy œ 3x# ey dx Ê ey dy œ 3x# dx Ê ' ey dy œ ' 3x# dx Ê ey œ x3 C Ê ey x3 œ C
Ê ey œ ex C Ê ey ex œ C
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
œ e.
586 13.
Chapter 9 Further Applications of Integration dy dx
sec# Èy Èy dy
œ Èy cos# Èy Ê dy œ ˆÈy cos# Èy‰ dx Ê
side, substitute u œ Èy Ê du œ
1 2Èy dy
Ê 2 du œ
1 Èy dy,
œ dx Ê '
sec# Èy Èy dy
œ ' dx. In the integral on the left-hand
and we have ' sec# u du œ ' dx Ê 2 tan u œ x C
Ê x 2 tan Èy œ C 14. È2xy dy dx œ 1 Ê dy œ Ê È2 15. Èx
dy dx
y3/2 3 2
dy œ
x1/2 " #
œ eyÈx Ê
1 È2xy
dx Ê È2Èydy œ
1 Èx
dx Ê È2 y1/2 dy œ x1/2 dx Ê È2 ' y1/2 dy œ ' x1/2 dx
3 C1 Ê È2 y3/2 œ 3Èx 32 C1 Ê È2 ˆÈy‰ 3Èx œ C, where C œ 32 C1
dy dx
œ
ey eÈ x Èx
Ê dy œ
ey eÈ x Èx dx
right-hand side, substitute u œ Èx Ê du œ
" #È x
eÈ x Ê ecy dy œ È dx Ê x
dx Ê 2 du œ
Ê ecy œ 2eu C1 Ê ecy œ 2eÈx C, where C œ C1
16. asec xb
dy dx
œ eysin x Ê
dy dx
" Èx
' ecy dy œ ' ÈeÈx dxÞ In the integral on the
dx, and we have
x
' ecy dy œ 2 ' eu du
œ eysin x cos x Ê dy œ aey esin x cos xbdx Ê ey dy œ esin x cos x dx
Ê ' ecy dy œ ' esin x cos x dx Ê ecy œ esin x C1 Ê ecy esin x œ C, where C œ C1
17.
dy dx
œ 2xÈ1 y2 Ê dy œ 2xÈ1 y2 dx Ê
dy È 1 y2
œ 2x dx Ê '
dy È 1 y2
œ ' 2x dx Ê sin" y œ x# C since kyk "
Ê y œ sinax2 Cb 18.
dy dx
œ
e2xcy exby 2y
Ê dy œ
e2xcy exby dx
Ê dy œ
e2x ecy ex ey dx
œ
ex e2y dx
Ê e2y dy œ ex dx Ê ' e2y dy œ ' ex dx Ê
Ê e 2ex œ C where C œ 2C1 19. y w œ x y Ê slope of 0 for the line y œ x. For x, y 0, y w œ x y Ê slope 0 in Quadrant I. For x, y 0, y w œ x y Ê slope 0 in Quadrant III. For kyk kxk, y 0, x 0, y w œ x y Ê slope 0 in Quadrant II above y œ x. For kyk kxk, y 0, x 0, y w œ x y Ê slope 0 in Quadrant II below y œ x. For kyk kxk, x 0, y 0, y w œ x y Ê slope 0 in Quadrant IV above y œ x. For kyk kxk, x 0, y 0, y w œ x y Ê slope 0 in Quadrant IV below y œ x. All of the conditions are seen in slope field (d). 20. y w œ y 1 Ê slope is constant for a given value of y, slope is 0 for y œ 1, slope is positive for y 1 and negative for y 1. These characteristics are evident in slope field (c).
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
e2y #
œ ex C1
Section 9.1 Slope Fields and Separable Differential Equations 21. y w œ xy Ê slope œ 1 on y œ x and 1 on y œ x. y w œ xy Ê slope œ 0 on the y-axis, excluding a0, 0b, and is undefined on the x-axis. Slopes are positive for x 0, y 0 and x 0, y 0 (Quadrants II and IV), otherwise negative. Field (a) is consistent with these conditions.
22. y w œ y2 x2 Ê slope is 0 for y œ x and for y œ x. For kyk kxk slope is positive and for kyk kxk slope is negative. Field (b) has these characteristics.
23.
24.
25-36. Example CAS commands: Maple: ode := diff( y(x), x ) = y(x); icA := [0, 1]; icB := [0, 2]; icC := [0,-1]; DEplot( ode, y(x), x=0..2, [icA,icB,icC], arrows=slim, linecolor=blue, title="#25 (Section 9.1)" ); Mathematica: To plot vector fields, you must begin by loading a graphics package. y: 0 STO > x: y (enter) y 0.1*(1 y2 ) STO > y: x 0.1 STO > x: y (enter, 10 times) The last value displayed gives yEuler a1b ¸ 1.3964 The exact solution: dy œ a1 y2 bdx Ê
dy 1 y2
œ dx Ê tan1 y œ x C; tan1 ya0b œ tan1 0 œ 0 œ 0 C Ê C œ 0
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
596
Chapter 9 Further Applications of Integration Ê tan1 y œ x Ê y œ tan x Ê yexact a"b œ tan 1 ¸ 1.5574
17. (a)
dy dx
œ 2y2 ax 1b Ê
dy y2
œ 2ax 1bdx Ê ' y2 dy œ ' a2x 2bdx Ê y" œ x2 2x C
Initial value: ya2b œ "# Ê 2 œ 22 2a2b C Ê C œ 2 1 Solution: y" œ x2 2x 2 or y œ x2 2x 2
ya3b œ 32 21a3b 2 œ 15 œ 0.2 (b) To find the approximation, set y1 œ 2y2 ax 1b and use EULERT with initial values x œ 2 and y œ "# and step size
0.2 for 5 Points. This gives ya3b ¸ 0.1851; error ¸ 0.0149. (c) Use step size 0.1 for 10 points. This gives ya3b ¸ 0.1929; error ¸ 0.0071. (d) Use step size 0.05 for 20 points. This gives ya3b ¸ 0.1965; error ¸ 0.0035. 18. (a)
dy dx
œy1Ê'
dy y1
œ ' dx Ê ln ky 1k œ x C Ê ky 1k œ exC Ê y 1 œ „ eC ex Ê y œ Aex 1
Initial value: ya0b œ 3 Ê 3 œ Ae0 1 Ê A œ 2 Solution: y œ 2ex 1 ya1b œ 2e 1 ¸ 6.4366 (b) To find the approximation, set y1 œ y 1 and use a graphing calculator or CAS with initial values x œ 0 and y œ 3 and step size 0.2 for 5 Points. This gives ya1b ¸ 5.9766; error ¸ 0.4599 (c) Use step size 0.1 for 10 points. This gives ya1b ¸ 6.1875; error ¸ 0.2491. (d) Use step size 0.05 for 20 points. This gives ya1b ¸ 6.3066; error ¸ 0.1300. 1 2 x2 2x 2 , so ya3b œ 0.2. To find the approximation, let zn œ yn1 2yn1 axn1 1bdx ay2n1 axn1 1b z2n ax2n 1bbdx with initial values x0 œ 2 and y0 œ "# . Use a spreadsheet, graphing
19. The exact solution is y œ yn œ y n 1
and
calculator, or CAS as indicated in parts (a) through (d). (a) Use dx œ 0.2 with 5 steps to obtain ya3b ¸ 0.2024 Ê error ¸ 0.0024. (b) Use dx œ 0.1 with 10 steps to obtain ya3b ¸ 0.2005 Ê error ¸ 0.0005. (c) Use dx œ 0.05 with 20 steps to obtain ya3b ¸ 0.2001 Ê error ¸ 0.0001. (d) Each time the step size is cut in half, the error is reduced to approximately one-fourth of what it was for the larger step size. 20. The exact solution is y œ 2ex 1, so ya1b œ 2e 1 ¸ 6.4366. To find the approximation, let zn œ yn1 ayn1 1bdx and yn œ yn1 ˆ ync1 2zn 2 ‰dx with initial value yn œ 3. Use a spreadsheet, graphing calculator, or CAS as indicated in parts (a) through (d). (a) Use dx œ 0.2 with 5 steps to obtain ya1b ¸ 6.4054 Ê error ¸ 0.0311. (b) Use dx œ 0.1 with 10 steps to obtain ya1b ¸ 6.4282 Ê error ¸ 0.0084 (c) Use dx œ 0.05 with 20 steps to obtain ya1b ¸ 6.4344 Ê error ¸ 0.0022 (d) Each time the step size is cut in half, the error is reduced to approximately one-fourth of what it was for the larger step size. 13-16. Example CAS commands: Maple: ode := diff( y(x), x ) = 2*x*exp(x^2);ic := y(0)=2; xstar := 1; dx := 0.1; approx := dsolve( {ode,ic}, y(x), numeric, method=classical[foreuler], stepsize=dx ): approx(xstar); exact := dsolve( {ode,ic}, y(x) ); eval( exact, x=xstar ); Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 9.3 Euler's Method evalf( % ); 17.
Example CAS commands: Maple: ode := diff( y(x), x ) = 2*y(x)*(x-1);ic := y(2)=-1/2; xstar := 3; exact := dsolve( {ode,ic}, y(x) ); # (a) eval( exact, x=xstar ); evalf( % ); approx1 := dsolve( {ode,ic}, y(x), # (b) numeric, method=classical[foreuler], stepsize=0.2 ): approx1(xstar); approx2 := dsolve( {ode,ic}, y(x), # (c) numeric, method=classical[foreuler], stepsize=0.1 ): approx2(xstar); approx3 := dsolve( {ode,ic}, y(x), # (d) numeric, method=classical[foreuler], stepsize=0.05 ): approx3(xstar);
19.
Example CAS commands: Maple: ode := diff( y(x), x ) = 2*y(x)*(x-1);ic := y(2)=-1/2; xstar := 3; approx1 := dsolve( {ode,ic}, y(x), # (a) numeric, method=classical[heunform], stepsize=0.2 ): approx1(xstar); approx2 := dsolve( {ode,ic}, y(x), # (b) numeric, method=classical[heunform], stepsize=0.1 ): approx2(xstar); approx3 := dsolve( {ode,ic}, y(x), # (c) numeric, method=classical[heunform], stepsize=0.05 ): approx3(xstar);
21.
Example CAS commands: Maple: ode := diff( y(x), x ) = x + y(x);ic := y(0)=-7/10; x0 := -4;x1 := 4;y0 := -4; y1 := 4; b := 1; P1 := DEplot( ode, y(x), x=x0..x1, y=y0..y1, arrows=thin, title="#21(a) (Section 9.3)" ): P1; Ygen := unapply( rhs(dsolve( ode, y(x) )), x,_C1 ); # (b) P2 := seq( plot( Ygen(x,c), x=x0..x1, y=y0..y1, color=blue ), c=-2..2 ): # (c) display( [P1,P2], title="#21(c) (Section 9.3)" ); CC := solve( Ygen(0,C)=rhs(ic), C ); # (d) Ypart := Ygen(x,CC); P3 := plot( Ypart, x=0..b, title="#21(d) (Section 9.3)" ): P3; euler4 := dsolve( {ode,ic}, numeric, method=classical[foreuler], stepsize=(x1-x0)/4 ): # (e) P4 := odeplot( euler4, [x,y(x)], x=0..b, numpoints=4, color=blue ):
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598
Chapter 9 Further Applications of Integration display( [P3,P4], title="#21(e) (Section 9.3)" ); euler8 := dsolve( {ode,ic}, numeric, method=classical[foreuler], stepsize=(x1-x0)/8 ): # (f) P5 := odeplot( euler8, [x,y(x)], x=0..b, numpoints=8, color=green ): euler16 := dsolve( {ode,ic}, numeric, method=classical[foreuler], stepsize=(x1-x0)/16 ): P6 := odeplot( euler16, [x,y(x)], x=0..b, numpoints=16, color=pink ): euler32 := dsolve( {ode,ic}, numeric, method=classical[foreuler], stepsize=(x1-x0)/32 ): P7 := odeplot( euler32, [x,y(x)], x=0..b, numpoints=32, color=cyan ): display( [P3,P4,P5,P6,P7], title="#21(f) (Section 9.3)" ); , # (g) < 4 | (x1-x0)/ 4 | evalf[5]( abs(1-eval(y(x),euler4(b))/eval(Ypart,x=b))*100 ) >, < 8 | (x1-x0)/ 8 | evalf[5]( abs(1-eval(y(x),euler8(b))/eval(Ypart,x=b))*100 ) >, < 16 | (x1-x0)/16 | evalf[5]( abs(1-eval(y(x),euler16(b))/eval(Ypart,x=b))*100 ) >, < 32 | (x1-x0)/32 | evalf[5]( abs(1-eval(y(x),euler32(b))/eval(Ypart,x=b))*100 ) > >;
13-24. Example CAS commands: Mathematica: (assigned functions, step sizes, and values for initial conditions may vary) For exercises 13 - 20, find the exact solution as follows. Set up two error lists. Clear[x, y, f] f[x_,y_]:= 2 y2 (x 1) a = 2; b = 1/2; xstar = 3; desol=DSolve[{y'[x] == f[x, y[x]], y[a] == b}, y[x], x] //Simplify actual[x_] = desol[[1, 1, 2]]; {xstar, actual[xstar]} errorlisteuler = { }; errorlisteulerimp = { }; pa = Plot[actual[x], {x, a, xstar}] Euler's method with error at x*. The Do command is used with a sequence of commands that are repeated n times. a = 2; b = -1/2; dx = 0.2; xstar = 3; n = (xstar a) /dx; solnslist = {{a,b}}; Do[ {new = b + f[a,b] dx, a = a + dx, b = new, AppendTo[solnslist, {a,b}]},{n}] solnslist error= actual[xstar] solnslist[[n, 2]] relativeerror= error / actual[xstar] AppendTo[errorlisteuler, error] pe = ListPlot[solnslist, PlotStyle Ä {Hue[.4], PointSize[0.02]}] Show[pa, pe] Rerun with different values for dx, starting from largest to smallest. After doing this, observe what happens to the error as the step size decreases by entering the input command: errorlisteuler Improved Euler's method. with error at x* a = 2; b = 1/2; dx = 0.2; xstar = 3; n = (xstar a) /dx; solnslist = {{a,b}}; Do[{new1 = b f[a,b] dx, new2 = b + (f[a, b] f[a+dx, new1])/2 dx, a = a dx, b = new2, AppendTo[solnslist, {a,b}]},{n}] solnslist error= actual[xstar] solnslist[[n, 2] Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 9.4 Graphical Solutions of Autonomous Differential Equations relativeerror= error / actual[xstar] AppendTo[errorlisteulerimp, error] peimp = ListPlot[solnslist, PlotStyle Ä {Hue[.8], PointSize[0.02]}] Show[pa, peimp] Rerun with different values for dx, starting from largest to smallest. After doing this, observe what happens to the error as the step size decreases by entering the input command: errorlisteulerimp You can also type Show[pa, pe, peimp]. This would be appropriate for a fixed value of dx with each method. You can also make a list of relative errors. Problems 21 - 24 involve use of code from section 9.1 together with the above code for Euler's method. 9.4 GRAPHICAL SOUTIONS OF AUTONOMOUS DIFFERENTIAL EQUATIONS 1. y w œ ay 2bay 3b (a) y œ 2 is a stable equilibrium value and y œ 3 is an unstable equilibrium. (b) yww œ a2y 1by w œ 2ay 2bˆy 12 ‰ay 3b
(c)
2. y w œ ay 2bay 2b (a) y œ 2 is a stable equilibrium value and y œ 2 is an unstable equilibrium. (b) yww œ 2yy w œ 2ay 2byay 2b
(c)
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599
600
Chapter 9 Further Applications of Integration
3. y w œ y3 y œ ay 1byay 1b (a) y œ 1 and y œ 1 is an unstable equilibrium and y œ 0 is a stable equilibrium value. (b) yww œ a3y2 1by w œ 3ay 1bŠy
1 È3 ‹yŠy
1 È3 ‹ay
1b
(c)
4. y w œ yay 2b (a) y œ 0 is a stable equilibrium value and y œ 2 is an unstable equilibrium. (b) yww œ a2y 2by w œ 2yay 1bay 2b
(c)
5. y w œ Èy, y 0 (a) There are no equilibrium values. 1 1 w Èy œ "# (b) yww œ 2È y y œ 2È y
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 9.4 Graphical Solutions of Autonomous Differential Equations (c)
6. y w œ y Èy, y 0 (a) y œ 1 is an unstable equilibrium. (b) yww œ Š1
1 2È y ‹
y w œ Š1
1 ˆ 2È y ‹ y
Èy‰ œ ˆÈy "# ‰ˆÈy 1‰
(c)
7. y w œ ay 1bay 2bay 3b (a) y œ 1 and y œ 3 is an unstable equilibrium and y œ 2 is a stable equilibrium value. (b) yww œ a3y2 12y 11bay 1bay 2bay 3b œ 3ay 1bŠy
6 È3 ‹ay 3
2bŠy
6 È3 3 ‹ay
(c)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
3b
601
602
Chapter 9 Further Applications of Integration
8. y w œ y3 y2 œ y2 ay 1b (a) y œ 0 and y œ 1 is an unstable equilibrium. (b) yww œ a3y2 2ybay3 y2 b œ y3 a3y 2bay 1b
(c)
9.
10.
dP dt
œ 1 2P has a stable equilibrium at P œ "# .
dP dt œ Pa1 2Pb has an unstable equilibrium d2 P dP dt2 œ a1 4Pb dt œ Pa1 4Pba1 2Pb
d2 P dt2
œ 2 dP dt œ 2a1 2Pb
at P œ 0 and a stable equilibrium at P œ "# .
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 9.4 Graphical Solutions of Autonomous Differential Equations
11.
12.
dP dt œ 2PaP 3b has a d2 P dP dt2 œ 2a2P 3b dt œ
dP dt d2 P dt2
stable equilibrium at P œ 0 and an unstable equilibrium at P œ 3. 4Pa2P 3baP 3b
œ 3Pa1 PbˆP "# ‰ has a stable equilibria at P œ 0 and P œ 1 an unstable equilibrium at P œ "# . 3 œ #3 a6P2 6P+1b dP dt œ # PŠP
3 È3 ˆ ‹ P 6
"# ‰ŠP
3 È3 ‹aP 6
1b
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603
604
Chapter 9 Further Applications of Integration
13.
Before the catastrophe, the population exhibits logistic growth and Patb Ä M0 , the stable equilibrium. After the catastrophe, the population declines logistically and Patb Ä M1 , the new stable equilibrium. 14.
dP dt
œ rPaM PbaP mb, r, M, m 0
The model has 3 equilibrium points. The rest point P œ 0, P œ M are asymptotically stable while P œ m is unstable. For initial populations greater than m, the model predicts P approaches M for large t. For initial populations less than m, the model predicts extinction. Points of inflection occur at P œ a and P œ b where a œ "3 M m ÈM2 mM m2 ‘ and b œ "3 M m ÈM2 mM m2 ‘. (a) The model is reasonable in the sense that if P m, then P Ä 0 as t Ä _; if m P M, then P Ä M as t Ä _; if P M, then P Ä M as t Ä _. (b) It is different if the population falls below m, for then P Ä 0 as t Ä _ (extinction). If is probably a more realistic model for that reason because we know some populations have become extinct after the population level became too low. (c) For P M we see that dP dt œ rPaM PbaP mb is negative. Thus the curve is everywhere decreasing. Moreover, P ´ M is a solution to the differential equation. Since the equation satisfies the existence and uniqueness conditions, solution trajectories cannot cross. Thus, P Ä M as t Ä _. (d) See the initial discussion above. (e) See the initial discussion above. 15.
dv dt
œg
k 2 mv ,
Equilibrium: Concavity:
g, k, m 0 and vatb 0
dv dt
d2 v dt2
œg
k 2 mv
œ 0 Ê v œ É mg k
ˆ k ‰ˆ œ 2ˆ mk v‰ dv dt œ 2 m v g
k 2‰ mv
(a)
(b)
160 (c) vterminal œ É 0.005 œ 178.9
ft s
œ 122 mph
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Section 9.4 Graphical Solutions of Autonomous Differential Equations
605
16. F œ Fp Fr ma œ mg kÈv dv kÈ v, va0b œ v0 dt œ g m Thus,
dv dt
‰ , the terminal velocity. If v0 ˆ mg ‰ , the object will fall faster and faster, approaching the œ 0 implies v œ ˆ mg k k 2
2
‰ , the object will slow down to the terminal velocity. terminal velocity; if v0 ˆ mg k 2
17. F œ Fp Fr ma œ 50 5kvk dv 1 dt œ m a50 5kvkb The maximum velocity occurs when
dv dt
œ 0 or v œ 10
ft sec .
18. (a) The model seems reasonable because the rate of spread of a piece of information, an innovation, or a cultural fad is proportional to the product of the number of individuals who have it (X) and those who do not (N X). When X is small, there are only a few individuals to spread the item so the rate of spread is slow. On the other hand, when (N X) is small the rate of spread will be slow because there are only a few indiciduals who can receive it during the interval of time. The rate of spread will be fastest when both X and (N X) are large because then there are a lot of individuals to spread the item and a lot of individuals to receive it. (b) There is a stable equilibrium at X œ N and an unstable equilibrium at X œ 0. d2 X dt2
dX 2 œ k dX dt aN Xb kX dt œ k XaN XbaN 2Xb Ê inflection points at X œ 0, X œ
N 2,
(c)
(d) The spread rate is most rapid when x œ
Eventually all of the people will receive the item.
œ VL RL i œ RL ˆ VR i‰, V, L, R 0 œ RL ˆ VR i‰ œ 0 Ê i œ VR
19. L di dt Ri œ V Ê
di dt d i 2 dt œ
Equilibrium: Concavity:
N 2.
2
di dt
ˆ R ‰2 ˆ VR i‰ ˆ RL ‰ di dt œ L
Phase Line:
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
and X œ N.
606
Chapter 9 Further Applications of Integration
If the switch is closed at t œ 0, then ia0b œ 0, and the graph of the solution looks like this:
As t Ä _, it Ä isteady state œ
V R.
(In the steady state condition, the self-inductance acts like a simple wire connector and, as
a result, the current throught the resistor can be calculated using the familiar version of Ohm's Law.) 20. (a) Free body diagram of he pearl:
(b) Use Newton's Second Law, summing forces in the direction of the acceleration: ˆ m m P ‰g mk v. mg Pg kv œ ma Ê dv dt œ (c) Equilibrium: Ê vterminal œ Concavity:
dv dt
œ
k amPbg mŠ k
v‹ œ 0
am Pbg k
d2 v dt2
ˆ k ‰ am k Pbg v‹ œ mk dv dt œ m Š 2
(d)
(e) The terminal velocity of the pearl is
am Pbg . k
9.5 APPLICATIONS OF FIRST ORDER DIFFERENTIAL EQUATIONS 1. Note that the total mass is 66 7 œ 73 kg, therefore, v œ v0 eakÎmbt Ê v œ 9e3.9tÎ73 3.9tÎ73 (a) satb œ ' 9e3.9tÎ73 dt œ 2190 C 13 e
Since sa0b œ 0 we have C œ
2190 13
3.9tÎ73 ‰ ˆ and lim satb œ lim 2190 œ 13 1 e tÄ_
tÄ_
2190 13
¸ 168.5
The cyclist will coast about 168.5 meters. 73 ln 9 (b) 1 œ 9e3.9tÎ73 Ê 3.9t 73 œ ln 9 Ê t œ 3.9 ¸ 41.13 sec It will take about 41.13 seconds. Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 9.5 Applications of First-Order Differential Equations 2. v œ v0 eakÎmbt Ê v œ 9ea59,000Î51,000,000bt Ê v œ 9e59tÎ51,000 (a) satb œ ' 9e59tÎ51,000 dt œ 459,0000 e59tÎ51,000 C 59 Since sa0b œ 0 we have C œ
459,0000 59
ˆ1 e59tÎ51,000 ‰ œ and lim satb œ lim 459,0000 59 tÄ_
tÄ_
459,0000 59
¸ 7780 m
The ship will coast about 7780 m, or 7.78 km. ln 9 59t (b) 1 œ 9e59tÎ51,000 Ê 51,000 œ ln 9 Ê t œ 51,000 ¸ 1899.3 sec 59 It will take about 31.65 minutes. 3. The total distance traveled œ v0km Ê a2.75bak39.92b œ 4.91 Ê k œ 22.36. Therefore, the distance traveled is given by the function satb œ 4.91ˆ1 ea22.36/39.92bt ‰. The graph shows satb and the data points.
4.
a0.80ba49.90b œ 1.32 Ê k œ 998 k 33 v0 m k 998 We know that k œ 1.32 and m œ 33a49.9b œ 20 . 33 v0 m ˆ ak/mbt ‰ Using Equation 3, we have: satb œ k 1 e œ 1.32ˆ1 v0 m k
5. (a)
œ coasting distance Ê
dP dt
œ 0.0015Pa150 Pb œ
0.255 150 Pa150
Pb œ
Thus, k œ 0.255 and M œ 150, and P œ Initial condition: Pa0b œ 6 Ê 6 œ Formula: P œ
150 0.255t 1 24ec0.255t Ê 1 24e ln 48 Ê t œ 0.255 ¸ 17.21 weeks 150 125 œ 1 24ec0.255t Ê 1 24e0.255t 120 Ê t œ ln0.255 ¸ 21.28
(b) 100 œ
M 1 Aeckt
150 1 Ae0
150 1 24ec0.255t
k M PaM
œ
e20t/33 ‰ ¸ 1.32a1 e0.606t b
Pb
150 1 Aec0.255t
Ê 1 A œ 25 Ê A œ 24
œ
3 2
Ê 24e0.255t œ
" #
Ê e0.255t œ
" 48
œ
6 5
Ê 24e0.255t œ
" 5
Ê e0.255t œ
" 120
Ê 0.255t œ ln 48 Ê 0.255t œ ln 120
It will take about 17.21 weeks to reach 100 guppies, and about 21.28 weeks to reach 125 guppies. 6. (a)
dP dt
œ 0.0004Pa250 Pb œ
0.1 250 Pa150
Pb œ
Thus, k œ 0.1 and M œ 250, and P œ
k M PaM Pb M œ 1 250 Aec0.1t 1 Aeckt
Initial condition: Pa0b œ 28, where t œ 0 represents the year 1970 250 111 28 œ 1 250 Ae0 Ê 28a1 Ab œ 250 Ê A œ 28 1 œ 14 ¸ 7.9286 Formula: P œ
250 c0.1t 1 111 14 e
or approximately P œ
250 1 7.9286ec0.1t
(b) The population Patb will round to 250 when Patb 249.5 Ê 249.5 œ Ê
a249.5bˆ111ec0.1t ‰ 14
œ 0.5 Ê e0.1t œ
14 55,389
Ê 0.1t œ ln
14 55,389
250 c0.1t 1 111 14 e
Ê 249.5ˆ1
Ê t œ 10 aln 55,389 ln 14b ¸ 82.8.
It will take about 83 years. 7. (a) Using the general solution form Example 2, part (c), dy dt
œ a0.08875 ‚ 107 ba8 ‚ 107 yby Ê yatb œ
M 1 AecrMt
œ
111 0.1t ‰ 14 e
8‚107 1 Aeca!Þ!))(&ba)bt
œ
8‚107 1 Aec0.71t
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
œ 250
607
608
Chapter 9 Further Applications of Integration Apply the initial condition: ya0b œ 1.6 ‚ 107 œ
(b) yatb œ 4 ‚ 107 œ
8‚107 1A
Ê
8‚107 1 4ec0.71t
8 1.6
8‚107 1 4ec0.71 ¸ 2.69671 ‚ lnˆ 1 ‰ 0.714 ¸ 1.95253 years.
1 œ 4 Ê ya1b œ
Ê 4e0.71t œ 1 Ê t œ
107 kg.
8. (a) If a part of the population leaves or is removed from the environment (e.g., a preserve or a region) each year, then c would represent the rate of reduction of the population due to this removal and/or migration. When grizzly bears become a nuisance (e.g., feeding on livestock) or threaten human safety, they are often relocated to other areas or even eliminated, but only after relocation efforts fail. In addition, bears are killed, sometimes accidently and sometimes maliciously. For an environment that has a capacity of about 100 bears, a realistic value for c would probably be between 0 and 4. (b)
Equilibrium solutions:
dP dt
œ 0 œ 0.001a100 PbP 1 Ê P2 100P 1000 œ 0 Ê Peq ¸ 11.27 (unstable) and
Peq ¸ 88.73 (stable) (c)
For 0 Pa0b Ÿ 11, the bear population will eventually disappear, for 12 Ÿ Pa0b Ÿ 88, the population will grow to about 89, the population will remain at about 89, and for Pa0b 89, the population will decrease to about 89 bears. 9. (a)
dy dt
œ 1 y Ê dy œ a1 ybdt Ê
dy 1y
œ dt Ê ln k1 yk œ t C1 Ê eln k1yk œ etC1 Ê k1 yk œ et eC1
1 y œ „ C2 et Ê y œ Cet 1, where C2 œ eC1 and C œ „ C2 . Apply the initial condition: ya0b œ 1 œ Ce0 1 Ê C œ 2 Ê y œ 2et 1. (b)
dy dt
œ 0.5a400 yby Ê dy œ 0.5a400 yby dt Ê
Example 2, part (c), we obtain Ê ' Š 1y ln¹ y cy400 ¹
Êe Ê
y y400
Êyœ
1 400 y ‹dy
1 1 400 Š y
1 400 y ‹dy
dy a400 yby
œ 0.5 dt. Using the partial fraction decomposition in
œ 0.5 dt Ê Š 1y
1 400 y ‹dy
œ 200 dt
œ ' 200 dt Ê lnkyk lnky 400k œ 200t C1 Ê ln¹ y y400 ¹ œ 200t C1
œ e200tC1 œ e200t eC1 Ê ¹ y y400 ¹ œ C2 e200t (where C2 œ eC1 ) Ê
y y 400
œ „ C2 e200t
œ Ce200t (where C œ „ C2 ) Ê y œ Ce200t y 400 Ce200t Ê a1 Ce200t by œ 400 Ce200t
400 Ce200t Ce200t 1
ya0b œ 2 œ
Êyœ
400 1 Ae0
400 1 C1 ec200t
œ
400 1 Aec200t ,
Ê A œ 199 Ê yatb œ
where A œ C1 . Apply the initial condition:
400 1 199ec200t
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Section 9.5 Applications of First-Order Differential Equations 10.
dP dt
œ raM PbP Ê dP œ raM PbP dt Ê
we obtain
1 ˆ1 M P
1 ‰ M P dP
dP aM PbP
œ r dt. Using the partial fraction decomposition in Example 6, part (c),
'
œ r dt Ê ˆ P1
Ê lnkPk lnkP Mk œ arMb Ê ¸ P P M ¸ œ C2 earMbt (where
'
1 ‰ ˆ P1 P 1 M ‰dP œ rM dt M P dP œ rM dt Ê P t C1 Ê ln¸ P P M ¸ œ arMb t C1 Ê eln¸ P c M ¸ œ earMbtC1 œ earMbt eC1 C2 œ eC1 ) Ê PPM œ „ C2 earMbt Ê PPM œ CearMbt (where C œ „ C2 )
Ê P œ CearMbt P M CearMbt Ê ˆ1 CearMbt ‰P œ M CearMbt Ê P œ ÊPœ 11. (a)
dP dt
M , 1 AecarMbt
œ kP2 Ê ' P2 dP œ ' k dt Ê P" œ kt C Ê P œ
Initial condition: Pa0b œ P0 Ê P0 œ C1 Ê C œ
dP dt
œ raM PbaP mb Ê
Ê
ÊPœ
M 1 C1 ecarMbt
aP 100ba1200 Pb dP a1200 PbaP 100b dt
1 P0
" kt C
P0 1 kP0 t
(b) There is a vertical asymptote at t œ 12. (a)
M CearMbt CearMbt 1
where A œ C1 .
Solution: P œ kt a11/P0 b œ
dP dt
1 kpO
œ ra1200 PbaP 100b Ê
œ 1100 r Ê
ˆ 12001 P
1 ‰ dP P 100 dt
1 dP a1200 PbaP 100b dt
œ 1100 r Ê
œrÊ
ˆ 12001 P
1100 dP a1200 PbaP 100b dt
1 ‰ P 100 dP
P 100 1200 P 1100 r t
where C œ „ eC1 Ê P 100 œ 1200Ce1100 r t CPe1100 r t Ê Pa1 Ce1100 r t b œ 1200Ce ÊPœ
1200Ce 100 Ce1100 r t 1
œ
c1100 r t 1200 100 C e 1 C1 ec1100 r t
(b) Apply the initial condition: 300 œ
1200 100Aec1100 r t 1 Aec1100 r t
ÊPœ
1200 100A 1A
œ 1100 r
œ 1100 r dt
Ê ' ˆ 12001 P P1100 ‰dP œ ' 1100 r dt Ê ln a1200 Pb ln aP 100b œ 1100 r t C1 P 100 ¸ P 100 C1 1100 r t ¸ P 100 ¸ Ê ln ¸ 1200 Ê P œ 1100 r t C1 Ê ln 1200 P œ 1100 r t C1 Ê 1200 P œ „ e e 1100 r t
œ Ce1100 r t
100
where A œ C1 .
Ê 300 300A œ 1200 100A Ê A œ
9 2
ÊPœ
2400 900Aec1100 r t Þ 2 9ec1100 r t
(Note that P Ä 1200 as t Ä _.) (c)
dP dt
œ raM PbaP mb Ê
Ê ˆ M 1 P
1 ‰ dP P m dt
1 dP aM PbaP mb dt
œrÊ
œ raM mb Ê ' ˆ M 1 P
Ê ln aM Pb ln aP mb œ aM mb r t Ê
Pm MP
Mm dP aM PbaP mb dt
Ê Pˆ1 Ce
aMmb r t
œ MCe
œ raM mb Ê
'
1 ‰ raM mbdt P m dP œ Pm ¸ ¸ C1 Ê ln M P œ aM mb r t aMmb r t aMmb r t
œ CeaMmb r t where C œ „ eC1 Ê P m œ MCe aMmb r t ‰
mÊPœ
aP mb aM Pb dP aM PbaP mb dt
C1 Ê
Pm MP
Apply the initial condition Pa0b œ P0 MmA 1 A
Ê P0 P0 A œ M mA Ê A œ
M P0 P0 m
ÊPœ
ÊPœ
caMcmb r t M m Ce 1 C1 ecaMcmb r t
ÊPœ
MaP0 mb maM P0 becaMcmb r t aP0 mb aM P0 becaMcmb r t
(Note that P Ä M as t Ä _ provided P0 m.) 13. y œ mx Ê
y x
orthogonals:
œmÊ dy dx
xy y x2 w
œ 0 Ê y w œ yx . So for
œ xy Ê y dy œ x dx Ê
y2 2
x2 2
œC
Ê x y œ C1 2
œ raM mb
œ „ eC1 eaMmb r t
CPe
MCeaMcmb r t m CeaMcmb r t 1
A œ C1 . P0 œ
609
2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
M mAecaMcmb r t 1 AecaMcmb r t
610
Chapter 9 Further Applications of Integration
14. y œ cx2 Ê Ê yw œ
y x2
2y x .
œcÊ
x2 y 2xy x4
œ 0 Ê x2 y w œ 2xy
w
So for the orthogonals:
dy dx
x œ 2y
2
Ê 2ydy œ xdx Ê y2 œ x2 C Ê y œ „ É x2 C, 2
C0
15. kx2 y2 œ 1 Ê 1 y2 œ kx2 Ê
1 y2 x2
œk
x a2yby ˆ1 y2 ‰2x Ê œ 0 Ê 2yx2 y w œ a1 y2 ba2xb x% ˆ1 y2 ‰a2xb ˆ1 y 2 ‰ Ê y w œ 2xy2 œ xy . So for the orthogonals: 2 ˆ1 y 2 ‰ 2 dy xy dy œ x dx Ê ln y y2 œ x2 C dx œ 1y2 Ê y 2
w
2x 16. 2x2 y2 œ c2 Ê 4x 2yy w œ 0 Ê y w œ 4x 2y œ y . For
orthogonals:
dy dx
œ
y 2x
Ê
œ
dy y
dx 2x
Ê ln y œ "# ln x C
Ê ln y œ ln x1/2 ln C1 Ê y œ C1 kxk1/2
17. y œ cex Ê
y ecx
œcÊ
ex y w c yaex bac1b aex b2
œ!
Ê ex y w œ yex Ê y w œ y. So for the orthogonals: dy dx
œ
1 y 2
Ê y dy œ dx Ê
y2 2
œxC
Ê y œ 2x C1 Ê y œ „ È2x C1
xŠ 1y ‹y c ln y w
18. y œ ekx Ê ln y œ kx Ê
ln y x
œkÊ
Ê Š xy ‹ y w ln y œ 0 Ê y w œ dy dx
y ln y x .
x y ln y Ê y ln y dy œ x dx 1 2 " 2 ˆ " 2‰ # y ln y 4 ay b œ # x 2 y2 ln y y2 œ x2 C1
x2
œ0
So for the orthogonals:
œ
Ê Ê
C
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 9.5 Applications of First-Order Differential Equations 2 w 19. 2x2 3y2 œ 5 and y2 œ x3 intersect at a1, 1b. Also, 2x2 3y2 œ 5 Ê 4x 6y y w œ 0 Ê y w œ 4x 6y Ê y a1, 1b œ 3
y21 œ x3 Ê 2y1 y1w œ 3x2 Ê y1w œ x2 2
20. (a) x dx y dy œ 0 Ê
y2 2 w
3x2 2y1
Ê y1w a1, 1b œ 32 . Since y w † y1w œ ˆ 23 ‰ˆ 32 ‰ œ 1, the curves are orthogonal.
œ C is the general equation
of the family with slope y œ xy . For the orthogonals: yw œ
y x
Ê
dy y
œ
dx x
Ê ln y œ ln x C or y œ C1 x
(where C1 œ e Ñ is the general equation of the orthogonals. C
(b) x dy 2y dx œ 0 Ê 2y dx œ x dy Ê Ê "# Š dy y ‹œ
dx x
dy 2y
œ
dx x
Ê "# ln y œ ln x C Ê y œ C1 x2 is
the equation for the solution family. " # ln
y ln x œ C Ê
"y # y
w
Ê slope of orthogonals is
1 x dy dx
œ 0 Ê yw œ
2y x
x œ 2y 2
Ê 2y dy œ x dx Ê y2 œ x2 C is the general equation of the orthogonals.
2". y2 œ 4a2 4ax and y2 œ 4b2 4bx Ê (at intersection) 4a2 4ax œ 4b2 4bx Ê a2 b2 œ xaa bb Ê aa bbaa bb œ aa bbx Ê x œ a b. Now, y2 œ 4a2 4aaa bb œ 4a2 4a2 4ab œ 4ab Ê y œ „ 2Èab. 4a Thus the intersections are at Ša b, „ 2Èab‹. So, y2 œ 4a2 4ax Ê y1w œ 2y which are equal to 4a and È 2Š2
4a 2Š2Èab‹ 4b 2Š2Èab‹
œ
È ba
and
È ba
at the intersections. Also, y œ 4b 4bx Ê 2
2
y2w
œ
4b 2y
which are equal to
œ É ba and É ba at the intersections. ay1w b † ay2w b œ ". Thus the curves are orthogonal.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
ab‹
4b 2Š2Èab‹
and
611
612
Chapter 9 Further Applications of Integration
CHAPTER 9 PRACTICE EXERCISES ".
dy dx
œ Èy cos2 Èy Ê
2. y w œ
3yax1b2 y 1
Ê
œ dx Ê 2tanÈy œ x C Ê y œ ˆtan1 ˆ x 2 C ‰‰
dy Èy cos2 Èy
ay 1 b y dy
3. yy w œ secay2 bsec2 x Ê
œ 3ax 1b2 dx Ê y ln y œ ax 1b3 C
y dy secay2 b
sinˆy2 ‰ 2
œ sec2 x dx Ê
œ tan x C Ê sinay2 b œ 2tan x C1
sin x 4. y cos2 axb dy sin x dx œ 0 Ê y dy œ cos 2 axb dx Ê
y2 2
œ cos1axb C Ê y œ „ É cosa2xb C1
2ax 2b3/2 a3x 4b 15 2ax 2b3/2 a3x 4b C“ 15
5. y w œ xey Èx 2 Ê ey dy œ xÈx 2 dx Ê ey œ 2ax 2b3/2 a3x 4b 15
Ê y œ ln’ 6. y w œ xyex Ê 2
dy y
2
C“ Ê y œ ln’
C Ê ey œ
2ax 2b3/2 a3x 4b 15
C
œ ex x dx Ê ln y œ "# ex C 2
2
7. sec x dy x cos2 y dx œ 0 Ê
dy cos2 y
x dx œ sec x Ê tan y œ cos x x sin x C
8. 2x2 dx 3Èy csc x dy œ 0 Ê 3Èy dy œ
2x2 csc x dx
Ê 2y3/2 œ 2a2 x2 bcos x 4x sin x C
Ê y3/2 œ a2 x2 bcos x 2x sin x C1 9. y w œ
ey xy
Ê yey dy œ
Ê ay 1bey œ ln kxk C
dx x
10. y w œ xexy csc y Ê y w œ
x ex ey csc
yÊ
ey csc y dy
œ x ex dx Ê
11. xax 1bdy y dx œ 0 Ê xax 1bdy œ y dx Ê
dy y
œ
ey 2 asin
dx x ax 1 b
y cos yb œ ax 1bex C
Ê ln y œ lnax 1b lnaxb C
Ê ln y œ lnax 1b lnaxb ln C1 Ê ln y œ lnŠ C1 axx 1b ‹ Ê y œ
12. y w œ ay2 1bax1 b Ê
dy
y 2 1
œ
Ê
dx x
1 lnŠ yy c b1‹
2
C1 ax 1b x
1 œ ln x C Ê lnŠ yy 1 ‹ œ 2ln x ln C1 Ê
y1 y1
œ C1 x2
13. 2y w y œ xex/2 Ê y w "# y œ x2 ex/2 .
' ˆ "‰ paxb œ "# , vaxb œ e c # dx œ ex/2 .
ex/2 y w "# ex/2 y œ ˆex/2 ‰ˆ x2 ‰ˆex/2 ‰ œ 14.
w
y 2
x 2
Ê
d ˆ x/2 dx e
y‰ œ
x 2
Ê ex/2 y œ
x2 4
2
C Ê y œ ex/2 Š x4 C‹
y œ ex sin x Ê y w 2y œ 2ex sin x.
paxb œ 2, vaxb œ e' 2dx œ e2x . e2x y w 2e2x y œ 2e2x ex sin x œ 2ex sin x Ê x
d 2x dx ae
yb œ 2ex sin x Ê e2x y œ ex asin x cos xb C
2x
Ê y œ e asin x cos xb Ce
15. xy w 2y œ 1 x1 Ê y w ˆ 2x ‰y œ vaxb œ e2'
dx x
1 x
1 x2 .
2
œ e2ln x œ eln x œ x2 .
x2 y w 2xy œ x 1 Ê
d 2 dx ax yb
œ x 1 Ê x2 y œ
x2 2
xCÊyœ
" #
1 x
C x2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 9 Practice Exercises
613
16. xy w y œ 2x ln x Ê y w ˆ 1x ‰y œ 2 ln x. vaxb œ e d ˆ1 dx x
' dxx
2 œ eln x œ 1x . ˆ 1x ‰y w ˆ 1x ‰ y œ 2x ln x Ê
† y‰ œ 2x ln x Ê
† y œ c ln x d2 C Ê y œ xc ln x d2 Cx
1 x
17. a1 ex bdy ayex ex bdx œ 0 Ê a1 ex by w ex y œ ex Ê y w œ ' a exdx b
vaxb œ e 1 b ex œ elnae 1b œ ex 1. x cx aex 1by w aex 1bˆ 1 e ex ‰y œ a1e ex b aex 1b Ê Êyœ 18.
dx dy
ecx C ex 1
ex 1 ex y
œ
ecx a1 e x b .
x
œ
e cx C
1by d œ ex Ê aex 1by œ ex C
d aex dx c
1 ex
x 4yey œ 0 Ê x w x œ 4yey . Let vayb œ e
' dy
œ ey . Then ey x w xey œ 4ye2y Ê
d y dy axe b
œ 4ye2y
Ê xey œ a2y 1be2y C Ê x œ a2y 1bey Cey 19. ax 3y2 b dy y dx œ 0 Ê x dy y dx œ 3y2 dy Ê
d dx axyb
œ 3y2 dy Ê xy œ y3 C '
20. y dx a3x y2 cos yb dx œ 0 Ê x w Š 3y ‹x œ y3 cos y. Let vayb œ e
3dy y
3
œ e3ln y œ eln y œ y3 .
Then y3 x w 3y2 x œ cos y and y3 x œ ' cos y dy œ sin y C. So x œ y3 asin y Cb 21.
œ exy2 Ê ey dy œ eax2b dx Ê ey œ eax2b C. We have ya0b œ 2, so e2 œ e2 C Ê C œ 2e2 and e œ eax2b 2e2 Ê y œ lnˆeax2b 2e2 ‰
22.
dy dx
dy dx y
œ
y ln y 1 x2
Ê etan
Ê
c1 a0bC
dy y ln y
œ
dx 1 x2
Ê lnaln yb œ tan1 axb C Ê y œ ee
So y w ax 1b2 Ê yax 1b2 œ
x x1.
' x b2 1 dx
Let vaxb œ e
2 ax 1 b a x
1b2 y œ
x3 3
C Ê y œ ax 1b2 Š x3
y œ ax 1b2 Š x3 3
x2 2 2
x 2
x ax 1 b a x
1b2 Ê
d dx yax
3
x2 2
dy dx
tanc1 axbbln 2
2
œ e2lnax1b œ elnax1b œ ax 1b2 .
2 1b2 ‘ œ xax 1b Ê yax 1b œ ' xax 1bdx
1‹ ' ˆ 2 ‰dx
25.
tanc1 a0bbC
C‹. We have ya0b œ 1 Ê 1 œ C. So
1 2 w ˆ2‰ 24. x dy dx 2y œ x 1 Ê y x y œ x x . Let vaxb œ e
So
. We have ya0b œ e2 Ê e2 œ ee
œ 2 Ê tan1 a0b C œ ln 2 Ê 0 C œ ln 2 Ê C œ ln 2 Ê y œ ee
w ˆ 2 ‰ 23. ax 1b dy dx 2y œ x Ê y x 1 y œ
Ê
tanc1 axbbC
d x4 2 3 2 dx ax yb œ x x Ê x y œ 4 2 4 2x2 1 y œ x4 4x1 2 "# œ x 4x 2
x2 2
CÊyœ
x2 4
x
C x2
œ eln x œ x2 . So x2 y w 2xy œ x3 x 2
"# . We have ya1b œ 1 Ê 1 œ
3x2 y œ x2 . Let vaxb œ e' 3x dx œ ex . So ex y w 3x2 ex y œ x2 ex Ê 2
3
3
3
3
We have ya0b œ 1 Ê e0 a1b œ 13 e0 C Ê 1 œ
3
1 3
d dx axyb
dy ˆy È y ‰
3
" #
3
Ê C œ 14 .
3
œ x2 ex Ê ex y œ 13 ex C.
3
cos x x .
C
3
4 3
Êyœ
1 3
43 ex
3
' 1 dx x œ eln x œ x.
Let vaxb œ e
œ cos x Ê xy œ ' cos x dx Ê xy œ sin x C. We have yˆ 12 ‰ œ 0 Ê ˆ 12 ‰0 œ 1 C
Ê C œ 1. So xy œ 1 sin x Ê y œ 27. x dy ˆy Èy‰dx œ 0 Ê
d x3 dx Še y‹
C Ê C œ 43 and ex y œ 13 ex
26. xdy ay cos xbdx œ 0 Ê xy w y cos x œ 0 Ê y w ˆ 1x ‰y œ So xy w xˆ 1x ‰y œ cos x Ê
3
1 4
œ
dx x
1 sin x x
Ê 2lnˆÈy 1‰ œ ln x C. We have ya1b œ 1 Ê 2 lnŠÈ1 1‹ œ ln 1 C
Ê 2 ln 2 œ C œ ln 22 œ ln 4. So 2 lnˆÈy 1‰ œ ln x ln 4 œ lna4xb Ê lnˆÈy 1‰ œ "# lna4xb œ lna4xb1/2 Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
614
Chapter 9 Further Applications of Integration Ê eln
ˆÈ y 1 ‰
28. y2 dx dy œ So
y3 3
1/2
œ elna4xb Ê Èy 1 œ 2Èx Ê y œ ˆ2Èx 1‰
ex e2x 1
Ê
e2x 1 ex dx
œ ex ex
1 3
œ
dy yc2
Ê
y3 3
2
œ ex ex C. We have ya0b œ 1 Ê
a1 b 3 3
œ e0 e0 C Ê C œ 13 .
Ê y3 œ 3aex ex b 1 Ê y œ c3aex ex b 1 d1/3 xc2
29. xy w ax 2by œ 3x3 ex Ê y w ˆ x x 2 ‰y œ 3x2 ex . Let vaxb œ e' ˆ x ‰dx œ ex2ln x œ xe2 . So x x ex w ex ˆ x 2 ‰ d ˆ y œ 3 Ê dx y † xe2 ‰ œ 3 Ê y † xe2 œ 3x C. We have ya1b œ 0 Ê 0 œ 3a1b C Ê C œ 3 x2 y x2 x Êy†
ex x2
œ 3x 3 Ê y œ x2 ex a3x 3b
30. y dx a3x xy 2bdy œ 0 Ê Payb œ
3 y
x
dx dy
3x xy 2 y
œ0Ê
dx dy
3x y
x œ 2y Ê
dx dy
Š 3y 1‹x œ 2y .
1 Ê ' Paybdy œ 3ln y y Ê vayb œ e3ln yy œ y3 ey
y3 ey x w y3 ey Š 3y 1‹x œ 2y2 ey Ê y3 ey x œ ' 2y2 ey dy œ 2ey ay2 2y 2b C Ê y3 œ Ê y3 œ
2ˆy2 2y 2‰ Cey . x 2 yb1 ˆ ‰ 2 y 2y 2 4e x
We have ya2b œ 1 Ê 1 œ
2a1 2 2b Cec1 2
Ê C œ 4e and
31. To find the approximate values let yn œ yn1 ayn1 cos xn1 ba0.1b with x0 œ 0, y0 œ 0, and 20 steps. Use a spreadsheet, graphing calculator, or CAS to obtain the values in the following table. x y x y 1.1 1.6241 0 0 1.2 1.8319 0.1 0.1000 1.3 2.0513 0.2 0.2095 1.4 2.2832 0.3 0.3285 1.5 2.5285 0.4 0.4568 1.6 2.7884 0.5 0.5946 1.7 3.0643 0.6 0.7418 1.8 3.3579 0.7 0.8986 1.9 3.6709 0.8 1.0649 2.0 4.0057 0.9 1.2411 1.0 1.4273 32. To find the approximate values let zn œ yn1 aa2 yn1 ba2 xn1 3bba0.1b and yn œ yn1 Š a2 ync1 ba2 xnc1 32b a2 zn ba2 xn 3b ‹a0.1b with initial values x0 œ 3, y0 œ 1, and 20 steps. Use a spreadsheet, graphing calculator, or CAS to obtain the values in the following table. x y x y 1.9 5.9686 3 1 1.8 6.5456 2.9 0.6680 1.7 6.9831 2.8 0.2599 1.6 7.2562 2.7 0.2294 1.5 7.3488 2.6 0.8011 1.4 7.2553 2.5 1.4509 1.3 6.9813 2.4 2.1687 1.2 6.5430 2.3 2.9374 1.1 5.9655 2.2 3.7333 1.0 5.2805 2.1 4.5268 2.0 5.2840
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 9 Practice Exercises 2ync1 " xnc1 2ync1 33. To estimate ya3b, let zn œ yn1 Š xncx1nc1 1 ‹a0.05b and yn œ yn1 # Š xnc1 1
xn 2zn xn 1 ‹a0.05b
615
with initial values
x0 œ 0, y0 œ 1, and 60 steps. Use a spreadsheet, graphing calculator, or CAS to obtain ya3b ¸ 0.9063. 34. To estimate ya4b, let zn œ yn1 Š
x2nc1 2ync1 1 ‹a0.05b xnc1
with initial values x0 œ 1, y0 œ 1, and 60 steps. Use a
spreadsheet, graphing calculator, or CAS to obtain ya4b ¸ 4.4974. 35. Let yn œ yn1 ˆ exnc1 b1ync1 b 2 ‰adxb with starting values x0 œ 0 and y0 œ 2, and steps of 0.1 and 0.1. Use a spreadsheet, programmable calculator, or CAS to generate the following graphs. (a)
(b) Note that we choose a small interval of x-values because the y-values decrease very rapidly and our calculator cannot handle the calculations for x Ÿ 1. (This occurs because the analytic solution is y œ 2 lna2 ex b, which has an asymptote at x œ ln 2 ¸ 0.69. Obviously, the Euler approximations are misleading for x Ÿ 0.7.)
y
y
36. Let zn œ yn1 Š eynncc11 xnncc11 ‹adxb and yn œ yn1 #" Š eynncc11 xnncc11 x2
x2
xn2 zn ezn xn ‹adxb
with starting values x0 œ 0 and y0 œ 0,
and steps of 0.1 and 0.1. Use a spreadsheet, programmable calculator, or CAS to generate the following graphs. (a) (b)
37.
x y dy dx
1 1.2 1.4 1.6 1 0.8 0.56 0.28
œ x Ê dy œ x dx Ê y œ
Ê 1 œ
" #
Ê ya2b œ
CÊCœ 2
2 2
3 2
œ
" #
32
x2 2
1.8 0.04
2.0 0.4
C; x œ 1 and y œ 1
Ê yaexactb œ
x2 2
3 2
is the exact value.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
616
Chapter 9 Further Applications of Integration
38.
x y œ
dy dx
1 1.2 1.4 1.6 1.8 2.0 1 0.8 0.6333 0.4904 0.3654 0.2544 1 x
Ê dy œ x1 dx Ê y œ lnkxk C; x œ 1 and y œ 1
Ê 1 œ ln 1 C Ê C œ 1 Ê yaexactb œ lnkxk 1 Ê ya2b œ ln 2 1 ¸ 0.3069 is the exact value.
39.
x y
1 1.2 1.4 1.6 1.8 2.0 1 1.2 0.488 1.9046 2.5141 3.4192
œ xy Ê
dy dx
Êyœe
dy y
x2 2 C
œ x dx Ê lnkyk œ x2 2
x2 2
C
x2 2
œ e † eC œ C1 e ; x œ 1 and y œ 1 x2
Ê 1 œ C1 e1/2 Ê C1 œ e1/2 yaexactb œ e1/2 † e 2 œ eˆx 1‰/2 Ê ya2b œ e3/2 ¸ 4.4817 is the exact value. 2
40.
x y
1 1.2 1.4 1.6 1.8 2.0 1 1.2 1.3667 1.5130 1.6452 1.7688
dy y2 1 dx œ y Ê y dy œ dx Ê 2 œ x " " 2 # œ 1 C Ê C œ # Ê y œ
C; x œ 1 and y œ 1
2x 1 Ê yaexactb œ È2x 1 Ê ya2b œ È3 ¸ 1.7321 is the exact value.
41.
dy dx
œ y2 1 Ê y w œ ay 1bay 1b. We have y w œ 0 Ê ay 1b œ 0, ay 1b œ 0 Ê y œ 1, 1.
(a) Equilibrium points are 1 (stable) and 1 (unstable) (b) y w œ y2 1 Ê y ww œ 2yy w Ê y ww œ 2yay2 1b œ 2yay 1bay 1b. So y ww œ 0 Ê y œ 0, y œ 1, y œ 1.
(c)
42.
dy dx
œ y y2 Ê y w œ ya1 yb. We have y w œ 0 Ê ya1 yb œ 0 Ê y œ 0, 1 y œ 0 Ê y œ 0, 1.
(a) The equilibrium points are 0 and 1. So, 0 is unstable and 1 is stable. (b) Let ïî œ increasing, íï œ decreasing. yw ! yw ! yw ! qqíïïïïïqqñqqïïïïïîqqñqqíïïïïïqqpy 0 1 y w œ y y2 Ê y ww œ y w 2yy w Ê y ww œ ay y2 b 2yay y2 b œ y y2 2y2 2y3 Ê y ww œ 2y3 3y2 y œ ya2y2 3y 1b Ê y ww œ ya2y 1bay 1b. So, y ww œ 0 Ê y œ 0, 2y 1 œ 0, y 1 œ 0 Ê y œ 0, y œ "# , Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 9 Additional and Advanced Exercises
617
y œ 1. Let ïî œ concave up, íï œ concave down. y ww ! y ww ! y ww ! y ww ! qíïïïïïqqñqqïïïïïîqqñqqíïïïïïqqñqqïïïïïîqpy 0 1 1/2 (c)
43. (a) Force œ Mass times Acceleration (Newton's Second Law) or F œ ma. Let a œ
dv dt
œ
dv ds
†
ds dt
œ v dv ds . Then
2 2 ma œ mgR2 s2 Ê a œ gR2 s2 Ê v dv Ê v dv œ gR2 s2 ds Ê ' v dv œ ' gR2 s2 ds ds œ gR s
Ê
v2 2
œ
ÊCœ
gR2 s C1 2 v0 2gR
Ê v2 œ Êv œ 2
(b) If v0 œ È2gR, then v2 œ
2gR2 s 2gR2 s
2gR s
2
2C1 œ
v20
2gR2 s
C. When t œ 0, v œ v0 and s œ R Ê v20 œ
2gR2 R
C
2gR 2
È2gR. Then Ê v œ É 2gR s , since v 0 if v0
ds dt
œ
È2gR2 Ès
Ê Ès ds œ È2gR2 dt
Ê ' s1/2 ds œ ' È2gR2 dt Ê 23 s3/2 œ È2gR2 t C1 Ê s3/2 œ ˆ 32 È2gR2 ‰t C; t œ 0 and s œ R Ê R3/2 œ ˆ 32 È2gR2 ‰a0b C Ê C œ R3/2 Ê s3/2 œ ˆ 32 È2gR2 ‰t R3/2 œ ˆ 32 RÈ2g‰t R3/2 3 œ R3/2 ˆ 32 R1/2 È2g‰t 1 ‘ œ R3/2 ’ Š
44.
v0 m k
a0.86ba30.84b k 0.8866t
œ coasting distance Ê
Ê satb œ 0.97a1 e
È2gR 2R ‹t
2/3 0‰ 0‰ ‘ ‘ Ê s œ R 1 ˆ 3v 1 “ œ R3/2 ˆ 3v 2R t 1 2R t
œ 0.97 Ê k ¸ 27.343. satb œ
v0 m ˆ k 1
eak/mbt ‰ Ê satb œ 0.97ˆ1 ea27.343/30.84bt ‰
b. A graph of the model is shown superimposed on a graph of the data.
CHAPTER 9 ADDITIONAL AND ADVANCED EXERCISES 1. (a)
dy dt
A œ kA V ac yb Ê dy œ k V ay cbdt Ê
dy yc
' œ k A V dt Ê
dy yc
A œ ' k A V dt Ê lnky ck œ k V t C1
Ê y c œ „ eC1 ek V t . Apply the initial condition, ya0b œ y0 Ê y0 œ c C Ê C œ y0 c A
Ê y œ c ay0 cbek V t . A (b) Steady state solution: y_ œ lim yatb œ lim c ay0 cbek V t ‘ œ c ay0 cba0b œ c A
tÄ_
tÄ_
2. Measure the amounts of oxygen involved in mL. Then the inflow of oxygen is 1000 mL/min (Assumed: it will take 5 minutes to deliver the 5L œ 5000mL); the amount of oxygen at t œ 0 is 210 mL; letting A œ the amount of oxygen in the flask, the concentration at time t is A mL/L; the outflow rate of oxygen is A mL/L (lb/sec). The rate of change in A, dA dt , equals the rate of gain (1000 mL/min) minus rate of loss (A mL/min). Thus: dA dA t dt œ 1000 A Ê 1000 A œ dt Ê lnaA 1000b œ t C Ê A 1000 œ Ce . At t œ 0, A œ 210, so C œ 790 and A œ 1000 790et . Thus, Aa5b œ 1000 790e5 ¸ 994.7 mL. The concentration is
994.7 mL 1000 mL
œ 99.47%.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
618
Chapter 9 Further Applications of Integration
3. The amount of CO2 in the room at time t is Aatb. The rate of change in the amount of CO2 ,
dA dt
is the rate of internal
production (R1 ) plus the inflow rate (R2 ) minus the outflow rate (R3 ). R1 œ ˆ20
breaths/min ‰ a30 student
R2 œ Š1000
ft3 CO2 ft3 min ‹Š0.0004 min ‹
A R3 œ Š 10,000 ‹1000 œ 0.1A dA dt
3
100 2 studentsbˆ 1728 ft3 ‰Š0.04 ft ftCO ‹ ¸ 1.39 3
œ 0.4
ft3 CO2 min
ft3 CO2 min
ft3 CO2 min
œ 1.39 0.4 0.1A œ 1.79 0.1A Ê Aw 0.1A œ 1.79. Let vatb œ e
' 0.1dt
. We have
' 0.1dt d ‹ dt ŠAe
' 0.1dt
œ 1.79e
Ê Ae0.1t œ ' 1.79e0.1t dt œ 17.9e0.1t C. At t œ 0, A œ a10,000ba0.0004b œ 4 ft3 CO2 Ê C œ 13.9 Ê A œ 17.9 13.9e0.1t . So Aa60b œ 17.9 13.9e0.1a60b ¸ 17.87 ft3 of CO2 in the 10,000 ft3 room. The percent of 17.87 CO2 is 10,000 ‚ 100 œ 0.18% 4.
damvb damvb dm dm dv dm dm dm dv dm dt œ F av ub dt Ê F œ dt av ub dt Ê F œ m dt v dt v dt u dt Ê F œ m dt u dt . dm dt œ b Ê m œ kbkt C. At t œ 0, m œ m0 , so C œ m0 and m œ m0 kbkt. u kb k m0 kbkt dv Thus, F œ am0 kbktb dv dt ukbk œ am0 kbktbkgk Ê dt œ g m0 kbkt Ê v œ gt u lnŠ m0 ‹ C1
v œ 0 at t œ 0 Ê C1 œ 0. So v œ gt u lnŠ m0 m0kbkt ‹ œ t œ 0 Ê y œ "# gt2 c’ t Š
m0 kbkt m0 kbkt kbk ‹ lnŠ m0 ‹
dy dt
Ê y œ ' ’ gt u lnŠ
m0 kbkt m0 ‹
“dt and u œ c, y œ 0 at
“
' 5. (a) Let y be any function such that vaxby œ ' vaxbQaxb dx C, vaxb œ e Paxb dx . Then ' Paxb dx ' d w w Ê v w axb œ œ e Paxb dx Paxb œ vaxbPaxb. dx avaxb † yb œ vaxb † y y † v axb œ vaxbQaxb. We have vaxb œ e
Thus vaxb † y w y † vaxb Paxb œ vaxbQaxb Ê y w y Paxb œ Qaxb Ê the given y is a solution.
(b) If v and Q are continuous on c a, b d and x − aa, bb, then Ê
d dx ’
'xx vatbQatb dt“ œ vaxbQaxb 0
'xx vatbQatb dt œ ' vaxbQaxb dx. So C œ y0 vax0 b ' vaxbQaxb dx. From part (a), vaxby œ ' vaxbQaxb dx C. 0
Substituting for C: vaxby œ ' vaxbQaxb dx y0 vax0 b ' vaxbQaxb dx Ê vaxby œ y0 vax0 b when x œ x0 . 6. (a) y w Paxby œ 0, yax0 b œ 0. Use vaxb œ e' Paxb dx as an integrating factor. Then
d dx avaxbyb
œ 0 Ê vaxby œ C
Ê y œ Ce' Paxb dx and y1 œ C1 e' Paxb dx , y2 œ C# e' Paxb dx , y1 ax0 b œ y2 ax0 b œ 0, y1 y2 œ aC1 C2 be' Paxb dx œ C3 e (b)
' Paxb dx
d y axb dx avaxbc 1
and y1 y2 œ 0 0 œ 0. So y1 y2 is a solution to y w Paxby œ 0 with yax0 b œ 0.
y2 axb db œ
' Paxb dx ' Paxb dx d e aC1 dx Še
C2 b ‘‹ œ
d dx aC1
C2 b œ
d dx aC3 b
œ !.
' dxd avaxbc y1 axb y2 axb dbdx œ avaxbc y1 axb y2 axb db œ ' ! dx œ C ' ' ' ' (c) y1 œ C1 e Paxb dx , y2 œ C# e Paxb dx , y œ y1 y2 . So yax0 b œ 0 Ê C1 e Paxb dx C# e Paxb dx œ ! Ê C1 C2 œ 0 Ê C1 œ C2 Ê y1 axb œ y2 axb for a x b.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
CHAPTER 10 CONIC SECTIONS AND POLAR COORDINATES 10.1 CONIC SECTIONS AND QUADRATIC EQUATIONS 1. x œ
y# 8
Ê 4p œ 8 Ê p œ 2; focus is (2ß 0), directrix is x œ 2 #
2. x œ y4 Ê 4p œ 4 Ê p œ 1; focus is (1ß 0), directrix is x œ 1 #
3. y œ x6 Ê 4p œ 6 Ê p œ 4. y œ
x# 2
Ê 4p œ 2 Ê p œ
1 #
3 #
; focus is ˆ!ß 3# ‰ , directrix is y œ
3 #
; focus is ˆ!ß 1# ‰ , directrix is y œ 1#
5.
x# 4
y# 9
œ 1 Ê c œ È4 9 œ È13 Ê foci are Š „ È13ß !‹ ; vertices are a „ 2ß 0b ; asymptotes are y œ „ 3# x
6.
x# 4
y# 9
œ 1 Ê c œ È9 4 œ È5 Ê foci are Š0ß „ È5‹ ; vertices are a0ß „ 3b
7.
x# 2
y# œ 1 Ê c œ È2 1 œ 1 Ê foci are a „ 1ß 0b ; vertices are Š „ È2ß !‹
8.
y# 4
x# œ 1 Ê c œ È4 1 œ È5 Ê foci are Š0ß „ È5‹ ; vertices are a!ß „ 2b ; asymptotes are y œ „ 2x
9. y# œ 12x Ê x œ
y# 1#
Ê 4p œ 12 Ê p œ 3;
focus is ($ß !), directrix is x œ 3
11. x# œ 8y Ê y œ
x# 8
Ê 4p œ 8 Ê p œ 2;
focus is (!ß 2), directrix is y œ 2
#
10. x# œ 6y Ê y œ x6 Ê 4p œ 6 Ê p œ focus is ˆ!ß 3# ‰ , directrix is y œ 3#
#
3 #
;
y 12. y# œ 2x Ê x œ # Ê 4p œ 2 Ê p œ " focus is ˆ # ß !‰ , directrix is x œ "#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
" #
;
620
Chapter 10 Conic Sections and Polar Coordinates
13. y œ 4x# Ê y œ
x# ˆ "4 ‰
" 4
Ê 4p œ
Ê pœ
" 16
#
14. y œ 8x# Ê y œ ˆx" ‰ Ê 4p œ
;
8
" ‰ " focus is ˆ!ß 16 , directrix is y œ 16
#
15. x œ 3y# Ê x œ ˆy" ‰ Ê 4p œ 3
focus is ˆ 1"# ß !‰ , directrix is x œ
#
#
" 3
" 1#
y 17. 16x# 25y# œ 400 Ê #x5 16 œ1 Ê c œ Èa# b# œ È25 16 œ 3
#
19. 2x# y# œ 2 Ê x# y# œ 1 Ê c œ Èa# b# œ È2 1 œ 1
" ‰ focus is ˆ!ß 32 , directrix is y œ
Ê pœ
" 1#
;
16. x œ 2y# Ê x œ
y# ˆ "# ‰
Ê 4p œ
" #
" 8
" 3#
Ê pœ
focus is ˆ 8" ß !‰ , directrix is x œ 8"
#
#
x 18. 7x# 16y# œ 112 Ê 16 y7 œ 1 Ê c œ Èa# b# œ È16 7 œ 3
#
#
20. 2x# y# œ 4 Ê x# y4 œ 1 Ê c œ Èa# b# œ È4 2 œ È2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Ê pœ
" 8
;
" 32
;
Section 10.1 Conic Sections and Quadratic Equations #
#
21. 3x# 2y# œ 6 Ê x# y3 œ 1 Ê c œ Èa# b# œ È3 2 œ 1
#
#
23. 6x# 9y# œ 54 Ê x9 y6 œ 1 Ê c œ Èa# b# œ È9 6 œ È3
#
#
x 22. 9x# 10y# œ 90 Ê 10 y9 œ 1 Ê c œ Èa# b# œ È10 9 œ 1
#
#
y x 24. 169x# 25y# œ 4225 Ê 25 169 œ1 Ê c œ Èa# b# œ È169 25 œ 12
#
25. Foci: Š „ È2ß !‹ , Vertices: a „ 2ß 0b Ê a œ 2, c œ È2 Ê b# œ a# c# œ 4 ŠÈ2‹ œ 2 Ê 26. Foci: a!ß „ 4b , Vertices: a0ß „ 5b Ê a œ 5, c œ 4 Ê b# œ 25 16 œ 9 Ê 27. x# y# œ 1 Ê c œ Èa# b# œ È1 1 œ È2 ; asymptotes are y œ „ x
x# 9
#
y# #5
œ1 #
x 28. 9x# 16y# œ 144 Ê 16 y9 œ 1 Ê c œ Èa# b# œ È16 9 œ 5; asymptotes are y œ „ 34 x
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
x# 4
y# #
œ1
621
622
Chapter 10 Conic Sections and Polar Coordinates
29. y# x# œ 8 Ê y8 x8 œ 1 Ê c œ Èa# b# œ È8 8 œ 4; asymptotes are y œ „ x
# # 30. y# x# œ 4 Ê y4 x4 œ 1 Ê c œ Èa# b# œ È4 4 œ 2È2; asymptotes are y œ „ x
31. 8x# 2y# œ 16 Ê x# y8 œ 1 Ê c œ Èa# b# œ È2 8 œ È10 ; asymptotes are y œ „ 2x
32. y# 3x# œ 3 Ê y3 x# œ 1 Ê c œ Èa# b# œ È3 1 œ 2; asymptotes are y œ „ È3x
# # 33. 8y# 2x# œ 16 Ê y# x8 œ 1 Ê c œ Èa# b# œ È2 8 œ È10 ; asymptotes are y œ „ x
y x 34. 64x# 36y# œ 2304 Ê 36 64 œ 1 Ê c œ Èa# b# œ È36 64 œ 10; asymptotes are y œ „ 4
#
#
#
#
#
#
#
#
3
35. Foci: Š!ß „ È2‹ , Asymptotes: y œ „ x Ê c œ È2 and
a b
œ 1 Ê a œ b Ê c# œ a# b# œ 2a# Ê 2 œ 2a#
Ê a œ 1 Ê b œ 1 Ê y# x# œ 1 36. Foci: a „ 2ß !b , Asymptotes: y œ „ Ê 4œ
4a# 3
" È3
x Ê c œ 2 and
Ê a# œ 3 Ê a œ È3 Ê b œ 1 Ê
x# 3
b a
œ
" È3
Ê bœ
a È3
4 3
Ê c# œ a# b# œ a#
y# œ 1
37. Vertices: a „ 3ß 0b , Asymptotes: y œ „ 43 x Ê a œ 3 and
b a
œ
4 3
Ê bœ
(3) œ 4 Ê
38. Vertices: a!ß „ 2b , Asymptotes: y œ „ 12 x Ê a œ 2 and
a b
œ
1 2
Ê b œ 2(2) œ 4 Ê
x# 9 y# 4
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
y# 16 x# 16
œ1 œ1
a# 3
œ
4a# 3
Section 10.1 Conic Sections and Quadratic Equations 39. (a) y# œ 8x Ê 4p œ 8 Ê p œ 2 Ê directrix is x œ 2, focus is (#ß !), and vertex is (!ß 0); therefore the new directrix is x œ 1, the new focus is (3ß 2), and the new vertex is (1ß 2)
40. (a) x# œ 4y Ê 4p œ 4 Ê p œ 1 Ê directrix is y œ 1, focus is (!ß 1), and vertex is (!ß 0); therefore the new directrix is y œ 4, the new focus is (1ß 2), and the new vertex is (1ß 3)
41. (a)
x# 16
y# 9
œ 1 Ê center is (!ß 0), vertices are (4ß 0)
and (%ß !); c œ Èa# b# œ È7 Ê foci are ŠÈ7ß 0‹ and ŠÈ7ß !‹ ; therefore the new center is (%ß $), the new vertices are (!ß 3) and (8ß 3), and the new foci are Š4 „ È7ß $‹
42. (a)
x# 9
y# 25
œ 1 Ê center is (!ß 0), vertices are (0ß 5) and (0ß 5); c œ Èa# b# œ È16 œ 4 Ê foci are (!ß 4) and (!ß 4) ; therefore the new center is (3ß 2), the new vertices are (3ß 3) and (3ß 7), and the new foci are (3ß 2) and (3ß 6)
43. (a)
x# 16
y# 9
œ 1 Ê center is (!ß 0), vertices are (4ß 0)
and (4ß 0), and the asymptotes are x4 œ „ y3 or Èa# b# œ È25 œ 5 Ê foci are y œ „ 3x 4 ;cœ (5ß 0) and (5ß 0) ; therefore the new center is (2ß 0), the new vertices are (2ß 0) and (6ß 0), the new foci are (3ß 0) and (7ß 0), and the new asymptotes are yœ „
3(x 2) 4
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623
624
Chapter 10 Conic Sections and Polar Coordinates
44. (a)
y# 4
x# 5
œ 1 Ê center is (!ß 0), vertices are (0ß 2)
and (0ß 2), and the asymptotes are yœ „
2x È5
y 2
œ „
x È5
or
; c œ Èa# b# œ È9 œ 3 Ê foci are
(0ß 3) and (0ß 3) ; therefore the new center is (0ß 2), the new vertices are (0ß 4) and (0ß 0), the new foci are (0ß 1) and (0ß 5), and the new asymptotes are 2x y2œ „ È 5 45. y# œ 4x Ê 4p œ 4 Ê p œ 1 Ê focus is ("ß 0), directrix is x œ 1, and vertex is (0ß 0); therefore the new vertex is (2ß 3), the new focus is (1ß 3), and the new directrix is x œ 3; the new equation is (y 3)# œ 4(x 2) 46. y# œ 12x Ê 4p œ 12 Ê p œ 3 Ê focus is (3ß 0), directrix is x œ 3, and vertex is (0ß 0); therefore the new vertex is (4ß 3), the new focus is (1ß 3), and the new directrix is x œ 7; the new equation is (y 3)# œ 12(x 4) 47. x# œ 8y Ê 4p œ 8 Ê p œ 2 Ê focus is (0ß 2), directrix is y œ 2, and vertex is (0ß 0); therefore the new vertex is (1ß 7), the new focus is (1ß 5), and the new directrix is y œ 9; the new equation is (x 1)# œ 8(y 7) Ê focus is ˆ!ß #3 ‰ , directrix is y œ 3# , and vertex is (0ß 0); therefore the new vertex is (3ß 2), the new focus is ˆ3ß "# ‰ , and the new directrix is y œ 7# ; the new equation is
48. x# œ 6y Ê 4p œ 6 Ê p œ
3 #
(x 3)# œ 6(y 2) 49.
x# 6
y# 9
œ 1 Ê center is (!ß 0), vertices are (0ß 3) and (!ß 3); c œ Èa# b# œ È9 6 œ È3 Ê foci are Š!ß È3‹
and Š!ß È3‹ ; therefore the new center is (#ß 1), the new vertices are (2ß 2) and (#ß 4), and the new foci are Š#ß 1 „ È3‹ ; the new equation is 50.
x# 2
(x 2)# 6
(y 1)# 9
œ1
y# œ 1 Ê center is (!ß 0), vertices are ŠÈ2ß !‹ and ŠÈ2ß !‹ ; c œ Èa# b# œ È2 1 œ 1 Ê foci are
(1ß 0) and ("ß !); therefore the new center is (3ß 4), the new vertices are Š3 „ È2ß 4‹ , and the new foci are (2ß 4) and (4ß 4); the new equation is 51.
x# 3
y# #
(x 3)# #
(y 4)# œ 1
œ 1 Ê center is (!ß 0), vertices are ŠÈ3ß !‹ and ŠÈ3ß !‹ ; c œ Èa# b# œ È3 2 œ 1 Ê foci are
(1ß 0) and ("ß !); therefore the new center is (2ß 3), the new vertices are Š2 „ È3ß 3‹ , and the new foci are (1ß 3) and (3ß 3); the new equation is 52.
x# 16
y# #5
(x 2)# 3
(y 3)# #
œ1
œ 1 Ê center is (!ß 0), vertices are (!ß &) and (!ß 5); c œ Èa# b# œ È25 16 œ 3 Ê foci are
(0ß 3) and (0ß 3); therefore the new center is (4ß 5), the new vertices are (4ß 0) and (4ß 10), and the new foci are (4ß 2) and (4ß 8); the new equation is 53.
x# 4
y# 5
(x 4)# 16
(y 5)# #5
œ1
œ 1 Ê center is (!ß 0), vertices are (2ß 0) and (2ß 0); c œ Èa# b# œ È4 5 œ 3 Ê foci are ($ß !) and
(3ß 0); the asymptotes are „
x #
œ
y È5
Ê yœ „
È5x #
; therefore the new center is (2ß 2), the new vertices are
(4ß 2) and (0ß 2), and the new foci are (5ß 2) and (1ß 2); the new asymptotes are y 2 œ „
È5 (x 2) #
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
; the new
Section 10.1 Conic Sections and Quadratic Equations equation is 54.
x# 16
y# 9
(x 2)# 4
(y 2)# 5
625
œ1
œ 1 Ê center is (!ß 0), vertices are (4ß 0) and (4ß 0); c œ Èa# b# œ È16 9 œ 5 Ê foci are (5ß !)
and (5ß 0); the asymptotes are „
x 4
œ
Ê yœ „
y 3
3x 4
; therefore the new center is (5ß 1), the new vertices are
(1ß 1) and (9ß 1), and the new foci are (10ß 1) and (0ß 1); the new asymptotes are y 1 œ „ the new equation is
(x 5)# 16
(y 1)# 9
3(x 5) 4
;
œ1
55. y# x# œ 1 Ê center is (!ß 0), vertices are (0ß 1) and (0ß 1); c œ Èa# b# œ È1 1 œ È2 Ê foci are Š!ß „ È2‹ ; the asymptotes are y œ „ x; therefore the new center is (1ß 1), the new vertices are (1ß 0) and (1ß 2), and the new foci are Š1ß 1 „ È2‹ ; the new asymptotes are y 1 œ „ (x 1); the new equation is (y 1)# (x 1)# œ 1 56.
y# 3
x# œ 1 Ê center is (!ß 0), vertices are Š0ß È3‹ and Š!ß È3‹ ; c œ Èa# b# œ È3 1 œ 2 Ê foci are (!ß #)
and (!ß 2); the asymptotes are „ x œ
y È3
Ê y œ „ È3x; therefore the new center is (1ß 3), the new vertices
are Š"ß $ „ È3‹ , and the new foci are ("ß &) and (1ß 1); the new asymptotes are y 3 œ „ È3 (x 1); the new equation is
(y 3)# 3
(x 1)# œ 1
57. x# 4x y# œ 12 Ê x# 4x 4 y# œ 12 4 Ê (x 2)# y# œ 16; this is a circle: center at C(2ß 0), a œ 4 58. 2x# 2y# 28x 12y 114 œ 0 Ê x# 14x 49 y# 6y 9 œ 57 49 9 Ê (x 7)# (y 3)# œ 1; this is a circle: center at C(7ß 3), a œ 1 59. x# 2x 4y 3 œ 0 Ê x# 2x 1 œ 4y 3 1 Ê (x 1)# œ 4(y 1); this is a parabola: V(1ß 1), F(1ß 0) 60. y# 4y 8x 12 œ 0 Ê y# 4y 4 œ 8x 12 4 Ê (y 2)# œ 8(x 2); this is a parabola: V(#ß 2), F(!ß #) 61. x# 5y# 4x œ 1 Ê x# 4x 4 5y# œ 5 Ê (x 2)# 5y# œ 5 Ê
(x 2)# 5
y# œ 1; this is an ellipse: the
center is (2ß 0), the vertices are Š2 „ È5ß 0‹ ; c œ Èa# b# œ È5 1 œ 2 Ê the foci are (4ß 0) and (!ß 0) #
62. 9x# 6y# 36y œ 0 Ê 9x# 6 ay# 6y 9b œ 54 Ê 9x# 6(y 3)# œ 54 Ê x6 (y 9 3) œ 1; this is an ellipse: the center is (0ß 3), the vertices are (!ß 0) and (!ß 6); c œ Èa# b# œ È9 6 œ È3 Ê the foci are Š0ß 3 „ È3‹ #
63. x# 2y# 2x 4y œ 1 Ê x# 2x 1 2 ay# 2y 1b œ 2 Ê (x 1)# 2(y 1)# œ 2 # Ê (x1) (y 1)# œ 1; this is an ellipse: the center is (1ß 1), the vertices are Š" „ È2ß "‹ ; 2
c œ Èa# b# œ È2 1 œ 1 Ê the foci are (2ß 1) and (0ß 1) 64. 4x# y# 8x 2y œ 1 Ê 4 ax# 2x 1b y# 2y 1 œ 4 Ê 4(x 1)# (y 1)# œ 4 Ê (x 1)#
(y1)# 4
œ 1; this is an ellipse: the center is (1ß 1), the vertices are (1ß 3) and
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626
Chapter 10 Conic Sections and Polar Coordinates
(1ß 1); c œ Èa# b# œ È4 1 œ È3 Ê the foci are Š1ß " „ È3‹ 65. x# y# 2x 4y œ 4 Ê x# 2x 1 ay# 4y 4b œ 1 Ê (x 1)# (y 2)# œ 1; this is a hyperbola: the center is (1ß 2), the vertices are (2ß 2) and (!ß 2); c œ Èa# b# œ È1 1 œ È2 Ê the foci are Š1 „ È2ß #‹ ; the asymptotes are y 2 œ „ (x 1) 66. x# y# 4x 6y œ 6 Ê x# 4x 4 ay# 6y 9b œ 1 Ê (x 2)# (y 3)# œ 1; this is a hyperbola: the center is (2ß 3), the vertices are (1ß 3) and (3ß 3); c œ Èa# b# œ È1 1 œ È2 Ê the foci are Š2 „ È2ß 3‹ ; the asymptotes are y 3 œ „ (x 2) 67. 2x# y# 6y œ 3 Ê 2x# ay# 6y 9b œ 6 Ê
(y 3)# 6
x# 3
œ 1; this is a hyperbola: the center is (!ß $),
the vertices are Š!ß 3 „ È6‹ ; c œ Èa# b# œ È6 3 œ 3 Ê the foci are (0ß 6) and (!ß 0); the asymptotes are y 3 È6
œ „
x È3
Ê y œ „ È2x 3
68. y# 4x# 16x œ 24 Ê y# 4 ax# 4x 4b œ 8 Ê
y# 8
(x 2)# 2
œ 1; this is a hyperbola: the center is (2ß 0),
the vertices are Š2ß „ È8‹ ; c œ Èa# b# œ È8 2 œ È10 Ê the foci are Š2ß „ È10‹ ; the asymptotes are y È8
œ „
x 2 È2
Ê y œ „ 2(x 2)
69.
70.
71.
72.
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Section 10.1 Conic Sections and Quadratic Equations
627
74. kx# y# k Ÿ 1 Ê 1 Ÿ x# y# Ÿ 1 Ê 1 Ÿ x# y# and x# y# Ÿ 1 Ê 1 y# x# and x# y# Ÿ 1
73.
75. Volume of the Parabolic Solid: V" œ '0 21x ˆh bÎ2
œ
1hb# 8
76. y œ '
; Volume of the Cone: V# œ w H
x dx œ
w H
#
Š x# ‹ C œ
wx# 2H
" 3
#
1 ˆ b# ‰ h œ
" 3
4h b#
x# ‰ dx œ 21h '0 Šx bÎ2
#
1 Š b4 ‹ h œ
1hb# 12
4x$ b# ‹
; therefore V" œ
C; y œ 0 when x œ 0 Ê 0 œ
w(0)# 2H
#
dx œ 21h ’ x2 3 #
bÎ2
x% b# “ !
V#
C Ê C œ 0; therefore y œ
wx# 2H
is the
equation of the cable's curve 77. A general equation of the circle is x# y# ax by c œ 0, so we will substitute the three given points into a c œ 1 Þ b c œ 1 ß Ê c œ 43 and a œ b œ 73 ; therefore this equation and solve the resulting system: 2a 2b c œ 8 à 3x# 3y# 7x 7y 4 œ 0 represents the circle 78. A general equation of the circle is x# y# ax by c œ 0, so we will substitute each of the three given points 2a 3b c œ 13 Þ into this equation and solve the resulting system:
3a 2b c œ 13 ß Ê a œ 2, b œ 2, and c œ 23; 4a 3b c œ 25 à
therefore x# y# 2x 2y 23 œ 0 represents the circle 79. r# œ (2 1)# (1 3)# œ 13 Ê (x 2)# (y 1)# œ 13 is an equation of the circle; the distance from the center to (1.1ß 2.8) is È(# 1.1)# (1 2.8)# œ È12.85 È13 , the radius Ê the point is inside the circle 80. (x 2)# (y 1)# œ 5 Ê 2(x 2) 2(y 1)
dy dx
œ0 Ê
dy dx
2 # # œ yx 1 ; y œ 0 Ê (x 2) (0 1) œ 5
Ê (x 2)# œ 4 Ê x œ 4 or x œ 0 Ê the circle crosses the x-axis at (4ß 0) and (!ß 0); x œ 0 Ê (0 2)# (y 1)# œ 5 Ê (y 1)# œ 1 Ê y œ 2 or y œ 0 Ê the circle crosses the y-axis at (!ß 2) and (!ß !). At (4ß 0): At (!ß !): At (!ß #):
dy dx dy dx dy dx
2 œ 40 1 œ 2 Ê the tangent line is y œ 2(x 4) or y œ 2x 8 2 œ 00 1 œ 2 Ê the tangent line is y œ 2x
2 œ 02 1 œ 2 Ê the tangent line is y 2 œ 2x or y œ 2x 2
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Chapter 10 Conic Sections and Polar Coordinates
81. (a) y# œ kx Ê x œ
y# k
; the volume of the solid formed by
Èkx
revolving R" about the y-axis is V" œ '0 œ
1 k#
Èkx
'0
y% dy œ
1x# Èkx 5
#
#
1 Š yk ‹ dy
; the volume of the right
circular cylinder formed by revolving PQ about the y-axis is V# œ 1x# Èkx Ê the volume of the solid formed by revolving R# about the y-axis is V$ œ V# V" œ
41x# Èkx 5
. Therefore we can see the
ratio of V$ to V" is 4:1.
(b) The volume of the solid formed by revolving R# about the x-axis is V" œ '0 1 ŠÈkt‹ dt œ 1k'0 t dt #
x
œ
1kx# #
x
. The volume of the right circular cylinder formed by revolving PS about the x-axis is #
V# œ 1 ŠÈkx‹ x œ 1kx# Ê the volume of the solid formed by revolving R" about the x-axis is 1kx# #
V$ œ V# V" œ 1kx#
œ
1kx# #
. Therefore the ratio of V$ to V" is 1:1.
82. Let P" (pß y" ) be any point on x œ p, and let P(xß y) be a point where a tangent intersects y# œ 4px. Now y# œ 4px Ê 2y
dy dx
œ 4p Ê
dy dx
œ
2p y
Ê y# yy" œ 2px 2p# . Since x œ Ê
" #
tangents from P" are m" œ
y# 4p
2p y" Èy#" 4p#
œ
dy dx
œ
#
y , we have y# yy" œ 2p Š 4p ‹ 2p# Ê y# yy" œ
2y" „ È4y#" 16p# #
y# yy" 2p# œ 0 Ê y œ
y y" x (p)
; then the slope of a tangent line from P" is
and m# œ
" #
2p y
y# 2p#
œ y" „ Èy#" 4p# . Therefore the slopes of the two 2p y" Èy#" 4p#
Ê m" m# œ
4p# y#" ay#" 4p# b
œ 1
Ê the lines are perpendicular 83. Let y œ É1
x# 4
on the interval 0 Ÿ x Ÿ 2. The area of the inscribed rectangle is given by
A(x) œ 2x Š2É1 Ê Aw (x) œ 4É1
x# 4‹ x# 4
œ 4xÉ1
x# É1 x4#
x# 4
(since the length is 2x and the height is 2y)
. Thus Aw (x) œ 0 Ê 4É1
x# 4
x# É1 x4#
œ 0 Ê 4 Š1
x# 4‹
x# œ 0 Ê x# œ 2
Ê x œ È2 (only the positive square root lies in the interval). Since A(0) œ A(2) œ 0 we have that A ŠÈ2‹ œ 4 is the maximum area when the length is 2È2 and the height is È2. 84. (a) Around the x-axis: 9x# 4y# œ 36 Ê y# œ 9 94 x# Ê y œ „ É9 94 x# and we use the positive root #
Ê V œ 2 '0 1 ŠÉ9 94 x# ‹ dx œ 2 '0 1 ˆ9 94 x# ‰ dx œ 21 9x 34 x$ ‘ ! œ 241 2
2
#
(b) Around the y-axis: 9x# 4y# œ 36 Ê x# œ 4 49 y# Ê x œ „ É4 49 y# and we use the positive root #
Ê V œ 2'0 1 ŠÉ4 49 y# ‹ dy œ 2 '0 1 ˆ4 49 y# ‰ dy œ 21 4y 3
85. 9x# 4y# œ 36 Ê y# œ œ
91 4
9x# 36 4
'24 ax# 4b dx œ 941 ’ x3
$
3
4 27
$
y$ ‘ ! œ 161
Ê y œ „ #3 Èx# 4 on the interval 2 Ÿ x Ÿ 4 Ê V œ '2 1 Š #3 Èx# 4‹ dx #
4
%
4x“ œ #
91 4
ˆ 64 ‰ ˆ8 ‰‘ œ 3 16 3 8
91 4
ˆ 56 ‰ 3 8 œ
31 4
(56 24) œ 241
86. x# y# œ 1 Ê x œ „ È1 y# on the interval 3 Ÿ y Ÿ 3 Ê V œ 'c3 1 ˆÈ1 y# ‰ dy œ 2'0 1 ˆÈ1 y# ‰ dy 3
œ 21'0 a1 y# b dy œ 21 ’y 3
$ y$ 3 “!
#
œ 241
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3
#
Section 10.1 Conic Sections and Quadratic Equations 87. Let y œ É16
x# on the interval 3 Ÿ x Ÿ 3. Since the plate is symmetric about the y-axis, x œ 0. For a
16 9
É16 aµ x ßµ y b œ xß #
vertical strip:
Ê mass œ dm œ $ dA œ $É16 # É16 16 9 x
µ y dm œ
#
Š$ É16
16 9
16 9
16 9
x#
, length œ É16
16 9
x# , width œ dx Ê area œ dA œ É16
16 9
x# dx
x# ‹ dx œ $ ˆ8 98 x# ‰ dx so the moment of the plate about the x-axis is
3
3
16 9
x# dx. Moment of the strip about the x-axis:
Mx œ ' µ y dm œ 'c3 $ ˆ8 89 x# ‰ dx œ $ 8x M œ 'c3 $ É16
629
8 27
$
x$ ‘ $ œ 32$ ; also the mass of the plate is
# x# dx œ 'c3 4$ É1 ˆ "3 x‰ dx œ 4$ 'c1 3È1 u# du where u œ 3
1
x 3
Ê 3 du œ dx; x œ 3
Ê u œ 1 and x œ 3 Ê u œ 1. Hence, 4$ 'c1 3È1 u# du œ 12$ 'c1 È1 u# du 1
œ 12$ ’ "2 ŠuÈ1 u# sin" u‹“ 88. y œ Èx# 1 Ê
dy dx
" #
œ
È2
1 ' œ É 2x x# 1 Ê S œ 0 #
–
89.
u œ È2x — Ä du œ È2 dx
drA dt
œ
drB dt
Ê
d dt
21 È2
ax# 1b
"
"
1
œ 61$ Ê y œ
"Î#
(2x) œ
x È x# 1
Mx M
œ
32$ 61$
#
œ
Ê Š dy dx ‹ œ
È2
16 31
. Therefore the center of mass is ˆ!ß 3161 ‰ .
x# x # 1
#
dy Ê Ê1 Š dx ‹ œ É1
È2
dy 1 È # ' 21yÊ1 Š dx ‹ dx œ '0 21Èx# 1 É 2x x# 1 dx œ 0 21 2x 1 dx ; #
'02 Èu# 1 du œ È21
#
#
’ " ŠuÈu# 1 ln Šu Èu# 1‹‹“ œ 2 2 !
1 È2
90. (a) tan " œ mL Ê tan " œ f w (x! ) where f(x) œ È4px ; œ
2p y!
" #
(4px)"Î# (4p) œ
(b) tan 9 œ mFP œ
œ
2p È4px
Ê f w (x! ) œ
2p È4px!
Ê tan " œ
(c) tan ! œ
’2È5 ln Š2 È5‹“
(rA rB ) œ 0 Ê rA rB œ C, a constant Ê the points P(t) lie on a hyperbola with foci at A
and B
f w (x) œ
x# x# 1
2p y! . y! 0 y! x! p œ x! p
tan 9 tan " 1 tan 9 tan "
y#! 2p(x! p) y! (x! p 2p)
œ
œ
y! 2p Šx p c y ‹
!
!
y! 2p 1 b Šx p‹ Šy ‹
!
4px! 2px! 2p# y! (x! p)
!
œ
2p(x! p) y! (x! p)
œ
2p y!
91. PF will always equal PB because the string has constant length AB œ FP PA œ AP PB. 92. (a) In the labeling of the accompanying figure we have y 1 œ tan t so the coordinates of A are (1ß tan t). The coordinates of P are therefore (1 rß tan t). Since 1# y# œ (OA)# , we have 1# tan# t œ (1 r)# Ê 1 r œ È1 tan# t œ sec t Ê r œ sec t 1. The coordinates of P are therefore (xß y) œ (sec tß tan t) Ê x# y# œ sec# t tan# t œ 1
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630
Chapter 10 Conic Sections and Polar Coordinates
(b) In the labeling of the accompany figure the coordinates of A are (cos tß sin t), the coordinates of C are (1ß tan t), and the coordinates of P are (1 dß tan t). By similar triangles,
d AB
œ
Ê
OC OA
d 1 cos t
œ
È1 tan# t 1
Ê d œ (1 cos t)(sec t) œ sec t 1. The coordinates of P are therefore (sec tß tan t) and P moves on the hyperbola x# y# œ 1 as in part (a).
93. x# œ 4py and y œ p Ê x# œ 4p# Ê x œ „ 2p. Therefore the line y œ p cuts the parabola at points (2pß p) and (2pß p), and these points are È[2p (2p)]# (p p)# œ 4p units apart. 94. x lim Š b x ba Èx# a# ‹ œ Ä_ a œ
b a x lim Ä_
’
x # ax # a # b “ x È x # a#
œ
b a x lim Ä_
b a x lim Ä_
’
Šx Èx# a# ‹ œ
a# “ x È x # a#
b a x lim Ä_
–
Šx Èx# a# ‹ Šx Èx# a# ‹ x È x # a#
œ0
10.2 CLASSIFYING CONIC SECTIONS BY ECCENTRICITY # y# 1. 16x# 25y# œ 400 Ê #x5 16 œ 1 Ê c œ Èa# b# œ È25 16 œ 3 Ê e œ ca œ 35 ; F a „ 3ß 0b ;
directrices are x œ 0 „
œ „
a e
5 ˆ 35 ‰
œ „
25 3
# x# 2. 7x# 16y# œ 112 Ê 16 y7 œ 1 Ê c œ Èa# b# œ È16 7 œ 3 Ê e œ ca œ 34 ; F a „ 3ß 0b ;
directrices are x œ 0 „
œ „
a e
4 ˆ 34 ‰
œ „
16 3
3. 2x# y# œ 2 Ê x# y2 œ 1 Ê c œ Èa# b# œ È2 1 œ 1 Ê e œ ca œ È12 ; F a0ß „ 1b ; #
directrices are y œ 0 „
a e
œ „
È2 Š È12 ‹
œ „2
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—
Section 10.2 Classifying Conic Sections by Eccentricity 4. 2x# y# œ 4 Ê
x# #
œ 1 Ê c œ Èa# b#
y# 4
œ È4 2 œ È2 Ê e œ directrices are y œ 0 „
a e
c a
œ
È2 2
; F Š0ß „ È2‹ ;
œ „ È22 œ „ 2È2 Š ‹ 2
# # 5. 3x# 2y# œ 6 Ê x# y3 œ 1 Ê c œ Èa# b# œ È3 2 œ 1 Ê e œ ca œ È13 ; F a0ß „ 1b ;
directrices are y œ 0 „
a e
œ „
È3
œ „3
Š È13 ‹
# x# 6. 9x# 10y# œ 90 Ê 10 y9 œ 1 Ê c œ Èa# b# œ È10 9 œ 1 Ê e œ ca œ È110 ; F a „ 1ß 0b ;
directrices are x œ 0 „
7. 6x# 9y# œ 54 Ê
x# 9
a e
œ „
y# 6
œ È9 6 œ È3 Ê e œ directrices are x œ 0 „
a e
È10 Š È110 ‹
œ „ 10
œ 1 Ê c œ Èa# b# c a
œ
È3 3
; F Š „ È3ß 0‹ ;
œ „ È33 œ „ 3È3 Š ‹ 3
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631
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Chapter 10 Conic Sections and Polar Coordinates
y# x# 8. 169x# 25y# œ 4225 Ê 25 169 œ 1 Ê c œ Èa# b# œ È169 25 œ 12 Ê e œ c œ 12 ; F a0ß „ 12b ; a
directrices are y œ 0 „
a e
œ „
13
13 ˆ 12 ‰ 13
œ „
169 12
x# #7
y# 36
9. Foci: a0ß „ 3b , e œ 0.5 Ê c œ 3 and a œ
c e
œ
3 0.5
œ 6 Ê b# œ 36 9 œ 27 Ê
10. Foci: a „ 8ß 0b , e œ 0.2 Ê c œ 8 and a œ
c e
œ
8 0.#
œ 40 Ê b# œ 1600 64 œ 1536 Ê
œ1 x# 1600
y# 1536
11. Vertices: a0ß „ 70b , e œ 0.1 Ê a œ 70 and c œ ae œ 70(0.1) œ 7 Ê b# œ 4900 49 œ 4851 Ê
œ1
x# 4851
y# 4900
œ1
Šx
9 È5 ‹
12. Vertices: a „ 10ß 0b , e œ 0.24 Ê a œ 10 and c œ ae œ 10(0.24) œ 2.4 Ê b# œ 100 5.76 œ 94.24 x# 100
Ê
y# 94.24
œ1
13. Focus: ŠÈ5ß !‹ , Directrix: x œ Ê eœ
È5 3
. Then PF œ
PF œ
Ê È(x x
256 ‰ 9
Ê
a e
œ
5 9
Šx#
16 3
x
Ê
81 5 ‹
Ê c œ ae œ 4 and 0)#
œ
x# y# œ
16 3
(y
" 4
18 È5
È3 #
¸x
Ê
x# ˆ 64 ‰ 3
4 9
a e
È5 3
16 3
Ê
œ
ae e#
#
œ
ae e#
¹x
Ê
16 3 #
Ê (x 4) y œ y# ˆ 16 ‰ 3
9 È5
9 È5 ¹
x# 9
x# y# œ 4 Ê
œ
16 ¸ 3
Ê
9 È5
PD Ê ÊŠx È5‹ (y 0)# œ
14. Focus: (%ß 0), Directrix: x œ 4)#
Ê c œ ae œ È5 and #
È5 3
Ê x# 2È5 x 5 y# œ
È œ #3 PD 3 ˆ # 32 4 x 3
9 È5
y# 4
4 e# 3 4
Ê
È5 e#
œ
Ê e# œ
9 È5 #
Ê Šx È5‹ y# œ
5 9
5 9
œ1 œ
16 3
ˆx
Ê e# œ
16 ‰# 3
Ê eœ
3 4
È3 #
. Then
#
Ê x 8x 16 y#
œ1
4 " # 15. Focus: (%ß 0), Directrix: x œ 16 Ê c œ ae œ 4 and ae œ 16 Ê ae e# œ 16 Ê e# œ 16 Ê e œ 4 Ê e œ PF œ 1 PD Ê È(x 4)# (y 0)# œ 1 kx 16k Ê (x 4)# y# œ 1 (x 16)# Ê x# 8x 16 y# #
œ
1 4
#
#
ax 32x 256b Ê
3 4
#
#
x y œ 48 Ê
œ
" #
1 È2
. Then PF œ #
1 È2
y# 48
. Then
œ1
#
PD Ê ÊŠx È2‹ (y 0)# œ
Šx 2È2‹ Ê x# 2È2 x 2 y# œ
1 #
4
x# 64
16. Focus: ŠÈ2ß !‹ , Directrix: x œ 2È2 Ê c œ ae œ È2 and Ê eœ
#
" #
a e
œ 2È 2 Ê
1 È2
ae e#
œ 2È 2 Ê
È2 e#
œ 2 È 2 Ê e# œ
#
¹x 2È2¹ Ê Šx È2‹ y#
Šx# 4È2 x 8‹ Ê
" #
x# y# œ 2 Ê
x# 4
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y# #
œ1
" #
Section 10.2 Classifying Conic Sections by Eccentricity 17. e œ
Ê take c œ 4 and a œ 5; c# œ a# b#
4 5
Ê 16 œ 25 b# Ê b# œ 9 Ê b œ 3; therefore x# #5
y# 9
œ1
18. The eccentricity e for Pluto is 0.25 Ê e œ
c a
œ 0.25 œ
" 4 #
Ê take c œ 1 and a œ 4; c# œ a# b# Ê 1 œ 16 b # # Ê b# œ 15 Ê b œ È15 ; therefore, x y œ 1 is a 16
15
model of Pluto's orbit.
19. One axis is from A("ß ") to B("ß 7) and is 6 units long; the other axis is from C($ß %) to D(1ß 4) and is 4 units long. Therefore a œ 3, b œ 2 and the major axis is vertical. The center is the point C("ß 4) and the ellipse is given by (x1)# 4
(y4)# 9
œ 1; c# œ a# b# œ 3# 2# œ 5
Ê c œ È5 ; therefore the foci are F Š1ß 4 „ È5‹ , the eccentricity is e œ yœ4„
a e
œ
c a
È5 3
, and the directrices are
œ 4 „ È5 œ 4 „ Š ‹ 3
9È 5 5
.
3
20. Using PF œ e † PD, we have È(x 4)# y# œ œ
4 9
ax# 18x 81b Ê
5 9
2 3
kx 9k Ê (x 4)# y# œ
x# y# œ 20 Ê 5x# 9y# œ 180 or
x# 36
#
y 20
4 9
(x 9)# Ê x# 8x 16 y#
œ 1.
21. The ellipse must pass through (!ß 0) Ê c œ 0; the point (1ß 2) lies on the ellipse Ê a 2b œ 8. The ellipse is tangent to the x-axis Ê its center is on the y-axis, so a œ 0 and b œ 4 Ê the equation is 4x# y# 4y œ 0. Next, 4x# y# 4y 4 œ 4 Ê 4x# (y 24)# œ 4 Ê x#
(y 2)# 4
standard symbols) Ê c# œ a# b# œ 4 1 œ 3 Ê c œ È3 Ê e œ
œ 1 Ê a œ 2 and b œ 1 (now using the c a
œ
È3 #
.
22. We first prove a result which we will use: let m" , and m# be two nonparallel, nonperpendicular lines. Let ! be the acute angle between the lines. Then tan ! œ 1m" m"mm## . To see this result, let )" be the angle of inclination of the line with slope m" , and )# be the angle of inclination of the line with slope m# . Assume m" m# . Then )" )# and we have ! œ )" )# . Then tan ! œ tan ()" )# ) )" tan )# m" m# œ 1tan tan )" tan )# œ 1 m" m# , since m" œ tan )" and and m# œ tan )# .
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Chapter 10 Conic Sections and Polar Coordinates
Now we prove the reflective property of ellipses (see the x# a#
accompanying figure): If # #
# #
# #
b x a y œ a b and y œ
b a
y# b#
œ 1, then Èa# x# Ê yw œ
bx aÈ a# x#
.
Let P(x! ß y! ) be any point on the ellipse Ê yw (x! ) œ
bx!
œ
aÉa# x#!
b # x ! a# y!
be the foci. Then mPF" œ
. Let F" (cß 0) and F# (cß 0)
y! x! c
and mPF# œ
y! x! c
. Let ! and
" be the angles between the tangent line and PF" and PF# , respectively. Then b# x
tan ! œ
! Œc a# y! c x! c ! y
b# x y Š1 c a# y (x! ! c) ‹ ! !
Similarly, tan " œ
b# cy!
œ
b# x#! b# x! c a# y#! a # y ! x ! a# y! c b# x! y!
œ
b# x! c ab# x#! a# y#! b a # y ! c aa # b # b x ! y!
È2 1
b # x! c a# b # a # y ! c c # x ! y !
œ
b# cy!
.
. Since tan ! œ tan " , and ! and " are both less than 90°, we have ! œ " .
23. x# y# œ 1 Ê c œ Èa# b# œ È1 1 œ È2 Ê e œ œ
œ
c a
œ È2 ; asymptotes are y œ „ x; F Š „ È2 ß !‹ ;
directrices are x œ 0 „
a e
œ „
" È2
# x# 24. 9x# 16y# œ 144 Ê 16 y9 œ 1 Ê c œ Èa# b# œ È16 9 œ 5 Ê e œ ca œ 54 ; asymptotes are
y œ „ 34 x; F a „ 5ß !b ; directrices are x œ 0 „ œ „
a e
"6 5
# # 25. y# x# œ 8 Ê y8 x8 œ 1 Ê c œ Èa# b# œ È8 8 œ 4 Ê e œ ca œ È48 œ È2 ; asymptotes are
y œ „ x; F a0ß „ 4b ; directrices are y œ 0 „ œ „
È8 È2
a e
œ „2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 10.2 Classifying Conic Sections by Eccentricity y# 4
26. y# x# œ 4 Ê
x# 4
œ 1 Ê c œ Èa# b#
œ È4 4 œ 2È2 Ê e œ
c a
œ
2È 2 2
œ È2 ; asymptotes
are y œ „ x; F Š0ß „ 2È2‹ ; directrices are y œ 0 „ œ „
2 È2
a e
œ „ È2
27. 8x# 2y# œ 16 Ê
x# 2
y# 8
œ È2 8 œ È10 Ê e œ
œ 1 Ê c œ Èa# b# c a
œ
È10 È2
œ È5 ; asymptotes
are y œ „ 2x; F Š „ È10ß !‹ ; directrices are x œ 0 „ œ „
È2 È5
635
œ „
a e
2 È10
# 28. y# 3x# œ 3 Ê y3 x# œ 1 Ê c œ Èa# b# œ È3 1 œ 2 Ê e œ ca œ È23 ; asymptotes are
y œ „ È3 x; F a0ß „ 2b ; directrices are y œ 0 „ œ „
È3 Š È23 ‹
œ „
a e
3 #
29. 8y# 2x# œ 16 Ê
y# 2
x# 8
œ È2 8 œ È10 Ê e œ
œ 1 Ê c œ Èa# b# c a
œ
È10 È2
œ È5 ; asymptotes
are y œ „ x# ; F Š0ß „ È10‹ ; directrices are y œ 0 „ œ „
È2 È5
œ „
a e
2 È10
y# x# 30. 64x# 36y# œ 2304 Ê 36 64 œ 1 Ê c œ Èa# b# 5 œ È36 64 œ 10 Ê e œ ca œ 10 6 œ 3 ; asymptotes are
y œ „ 43 x; F a „ 10ß !b ; directrices are x œ 0 „ œ „
6 ˆ 53 ‰
œ „
a e
18 5
31. Vertices a!ß „ 1b and e œ 3 Ê a œ 1 and e œ
c a
œ 3 Ê c œ 3a œ 3 Ê b# œ c# a# œ 9 1 œ 8 Ê y#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
x# 8
œ1
636
Chapter 10 Conic Sections and Polar Coordinates
y# 1#
œ1
œ 3 Ê c œ 3a Ê a œ 1 Ê b# œ c# a# œ 9 1 œ 8 Ê x#
y# 8
œ1
32. Vertices a „ 2ß !b and e œ 2 Ê a œ 2 and e œ 33. Foci a „ 3ß !b and e œ 3 Ê c œ 3 and e œ
c a
34. Foci a!ß „ 5b and e œ 1.25 Ê c œ 5 and e œ œ 25 16 œ 9 Ê
#
y 16
#
x 9
œ 2 Ê c œ 2a œ 4 Ê b# œ c# a# œ 16 4 œ 12 Ê
c a
c a
œ 1.25 œ
5 4
Ê cœ
a Ê 5œ
5 4
5 4
x# 4
a Ê a œ 4 Ê b# œ c# a#
œ1
4 # È2 . Then 35. Focus (4ß 0) and Directrix x œ 2 Ê c œ ae œ 4 and ae œ 2 Ê ae e# œ 2 Ê e# œ 2 Ê e œ # Ê e œ PF œ È2 PD Ê È(x 4)# (y 0)# œ È2 kx 2k Ê (x 4)# y# œ 2(x 2)# Ê x# 8x 16 y#
œ 2 ax# 4x 4b Ê x# y# œ 8 Ê
x# 8
y# 8
œ1
36. Focus ŠÈ10ß !‹ and Directrix x œ È2 Ê c œ ae œ È10 and
a e
œ È2 Ê
ae e#
œ È2 Ê
È10 e#
œ È 2 Ê e# œ È 5
#
#
Ê e œ %È5 . Then PF œ %È5 PD Ê ÊŠx È10‹ (y 0)# œ %È5 ¹x È2¹ Ê Šx È10‹ y# #
œ È5 Šx È2‹ Ê x# 2È10 x 10 y# œ È5 Šx# 2È2 x 2‹ Ê Š1 È5‹ x# y# œ 2È5 10 Ê
Š1 È5‹ x# #È5 10
y# 2È5 10
œ1 Ê
x# 2È 5
y# 10 2È5
œ1
37. Focus (2ß 0) and Directrix x œ "# Ê c œ ae œ 2 and
a e
œ
" # #
Ê
ae e#
œ
" #
Ê
PF œ 2PD Ê È(x 2)# (y 0)# œ 2 ¸x "# ¸ Ê (x 2) y# œ 4 ˆx œ 4 ˆx# x "4 ‰ Ê 3x# y# œ 3 Ê x#
#
y 3
2 e# " ‰# #
œ
" #
Ê e# œ 4 Ê e œ 2. Then
Ê x# 4x 4 y#
œ1
6 # È3. Then 38. Focus (6ß 0) and Directrix x œ # Ê c œ ae œ 6 and ae œ # Ê ae e# œ # Ê e# œ # Ê e œ 3 Ê e œ PF œ È3 PD Ê È(x 6)# (y 0)# œ È3 kx 2k Ê (x 6)# y# œ 3(x 2)# Ê x# 12x 36 y# x# 1#
œ 3 ax# 4x 4b Ê 2x# y# œ 24 Ê 39. È(x 1)# (y 3)# œ
3 #
y# 24
œ1
ky 2k Ê x# 2x 1 y# 6y 9 œ
Ê 4 ax# 2x 1b 5 ay# 12y 36b œ 4 4 180 Ê 40. c# œ a# b# Ê b# œ c# a# ; e œ x# a#
#
y b#
œ" Ê
x# a#
#
y a # ae # 1 b
c a
(y6)# 36
9 4
ay# 4y 4b Ê 4x# 5y# 8x 60y 4 œ 0
(x1)# 45
œ1
Ê c œ ea Ê c# œ e# a# Ê b# œ e# a# a# œ a# ae# 1b ; thus,
œ 1; the asymptotes of this hyperbola are y œ „ ae# 1bx Ê as e increases, the
absolute values of the slopes of the asymptotes increase and the hyperbola approaches a straight line. 41. To prove the reflective property for hyperbolas: x# a#
y# b#
œ 1 Ê a# y# œ b# x# a# b# and
dy dx
œ
xb# ya#
.
Let P(x! ß y! ) be a point of tangency (see the accompanying figure). The slope from P to F(cß 0) is x!y! c and from P to F# (cß 0) it is
y! x ! c
. Let the tangent through P meet
the x-axis in point A, and define the angles nF" PA œ ! and nF# PA œ " . We will show that tan ! œ tan " . From the preliminary result in Exercise 22,
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 10.3 Quadratic Equations and Rotations x b#
tan ! œ
! Πy! a# c x! c !
x b# y! 1 b Š y! a# ‹ Š x ! c‹ ! y
tan " œ
y
! Πx! c
x! b# y! a#
y! x! b# 1 Šx c ‹ Š y a# ‹
!
!
œ
œ
x#! b# x! b# cy#! a# x ! y ! a # y ! a # c x! y! b #
b# y! c
œ
a# b# x! b# c x ! y ! c # y! a# c
œ
b# y! c
. In a similar manner,
. Since tan ! œ tan " , and ! and " are acute angles, we have ! œ " .
42. From the accompanying figure, a ray of light emanating from the focus A that met the parabola at P would be reflected from the hyperbola as if it came directly from B (Exercise 41). The same light ray would be reflected off the ellipse to pass through B. Thus BPC is a straight line. Let " be the angle of incidence of the light ray on the hyperbola. Let ! be the angle of incidence of the light ray on the ellipse. Note that ! " is the angle between the tangent lines to the ellipse and hyperbola at P. Since BPC is a straight line, 2! 2" œ 180°. Thus ! " œ 90°. 10.3 QUADRATIC EQUATIONS AND ROTATIONS 1. x# 3xy y# x œ 0 Ê B# 4AC œ (3)# 4(1)(1) œ 5 0 Ê Hyperbola 2. 3x# 18xy 27y# 5x 7y œ 4 Ê B# 4AC œ (18)# 4(3)(27) œ 0 Ê Parabola 3. 3x# 7xy È17y# œ 1 Ê B# 4AC œ (7)# 4(3) È17 ¸ 0.477 0 Ê Ellipse #
4. 2x# È15 xy 2y# x y œ 0 Ê B# 4AC œ ŠÈ15‹ 4(2)(2) œ 1 0 Ê Ellipse 5. x# 2xy y# 2x y 2 œ 0 Ê B# 4AC œ 2# 4(1)(1) œ 0 Ê Parabola 6. 2x# y# 4xy 2x 3y œ 6 Ê B# 4AC œ 4# 4(2)(1) œ 24 0 Ê Hyperbola 7. x# 4xy 4y# 3x œ 6 Ê B# 4AC œ 4# 4(1)(4) œ 0 Ê Parabola 8. x# y# 3x 2y œ 10 Ê B# 4AC œ 0# 4(1)(1) œ 4 0 Ê Ellipse (circle) 9. xy y# 3x œ 5 Ê B# 4AC œ 1# 4(0)(1) œ 1 0 Ê Hyperbola 10. 3x# 6xy 3y# 4x 5y œ 12 Ê B# 4AC œ 6# 4(3)(3) œ 0 Ê Parabola 11. 3x# 5xy 2y# 7x 14y œ 1 Ê B# 4AC œ (5)# 4(3)(2) œ 1 0 Ê Hyperbola 12. 2x# 4.9xy 3y# 4x œ 7 Ê B# 4AC œ (4.9)# 4(2)(3) œ 0.01 0 Ê Hyperbola 13. x# 3xy 3y# 6y œ 7 Ê B# 4AC œ (3)# 4(1)(3) œ 3 0 Ê Ellipse 14. 25x# 21xy 4y# 350x œ 0 Ê B# 4AC œ 21# 4(25)(4) œ 41 0 Ê Hyperbola 15. 6x# 3xy 2y# 17y 2 œ 0 Ê B# 4AC œ 3# 4(6)(2) œ 39 0 Ê Ellipse
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
637
638
Chapter 10 Conic Sections and Polar Coordinates
16. 3x# 12xy 12y# 435x 9y 72 œ 0 Ê B# 4AC œ 12# 4(3)(12) œ 0 Ê Parabola 1 1 # Ê !œ 4 ; È È xw sin ! yw cos ! Ê x œ xw #2 yw #2 , y È È È È Š #2 xw #2 yw ‹ Š #2 xw #2 yw ‹ œ 2 Ê "#
17. cot 2! œ yœ Ê
18. cot 2! œ
AC B
AC B
œ
11 1
œ
therefore x œ xw cos ! yw sin !,
œ 0 Ê 2! œ
0 1
œ 0 Ê 2! œ
1 #
Ê !œ
È2 #
œ xw
xw # "# yw # œ 2 Ê xw # yw # œ 4 Ê Hyperbola 1 4
; therefore x œ xw cos ! yw sin !,
È2 w È2 w È2 # y # ,yœ x # È È È È Š #2 xw #2 yw ‹ Š #2 xw #2 yw ‹
y œ xw sin ! yw cos ! Ê x œ xw Ê Š Ê
" #
È2 #
xw
#
È2 #
yw ‹
È2 #
yw
yw
È2 # È Š #2
xw
xw # xw yw "# yw # "# xw # "# yw # "# xw # xw yw "# yw # œ 1 Ê
19. cot 2! œ
AC B
31 2È 3
œ
œ
" È3
1 3
Ê 2! œ
1 6
Ê !œ
È2 #
#
yw ‹ œ 1
xw # "# yw # œ 1 Ê 3xw # yw # œ 2 Ê Ellipse
3 #
; therefore x œ xw cos ! yw sin !,
È3 w È3 w " w 1 w # x # y,yœ # x # y È È Š #3 xw 1# yw ‹ Š 1# xw #3 yw ‹
y œ xw sin ! yw cos ! Ê x œ Ê 3Š
È3 #
#
xw 1# yw ‹ 2È3
8È3 Š "# xw 20. cot 2! œ
AC B
È3 #
œ
È3 #
È3 #
#
yw ‹ 8 Š
È3 #
xw "# yw ‹
yw ‹ œ 0 Ê 4xw # 16yw œ 0 Ê Parabola
12 È 3
œ
" È3
#
1 3
Ê 2! œ
y œ xw sin ! yw cos ! Ê x œ Ê Š
Š 1# xw
xw 1# yw ‹ È3 Š
È3 #
È3 #
1 6
Ê !œ
xw #1 yw , y œ
" #
; therefore x œ xw cos ! yw sin !,
xw È3 #
xw 1# yw ‹ Š 1# xw
È3 #
yw
yw ‹ 2 Š 1# xw
È3 #
#
yw ‹ œ 1 Ê
" #
xw # 5# yw # œ 1
Ê xw # 5yw # œ 2 Ê Ellipse 21. cot 2! œ
AC B
œ
11 2
œ 0 Ê 2! œ
y œ xw sin ! yw cos ! Ê x œ Ê Š
È2 #
xw
#
È2 #
yw ‹ 2 Š
È2 #
È2 #
1 2
Ê !œ È2 #
; therefore x œ xw cos ! yw sin !,
È2 w È2 w # x # y È2 w È2 w È2 w È2 # y ‹Š # x # y ‹ Š #
xw
xw
1 4
yw , y œ
xw
È2 #
#
yw ‹ œ 2 Ê yw # œ 1
Ê Parallel horizontal lines 22. cot 2! œ
AC B
œ
31 2 È 3
œ È"3 Ê 2! œ
y œ xw sin ! yw cos ! Ê x œ Ê 3 Š 1# xw
È3 #
#
1 #
xw
yw ‹ 2È3 Š 1# xw
È3 # È3 #
21 3
Ê !œ
yw , y œ yw ‹ Š
È3 #
È3 #
1 3
; therefore x œ xw cos ! yw sin !,
xw 1# yw
xw 1# yw ‹ Š
È3 #
#
xw 1# yw ‹ œ 1 Ê 4yw # œ 1
Ê Parallel horizontal lines 23. cot 2! œ
AC B
œ
È2 È2 2È 2
y œ xw sin ! yw cos ! Ê x œ È Ê È 2 Š # 2 xw
È2 #
Ê !œ
#
È2 #
xw
È2 #
yw , y œ
È2 #
xw
È2 #
yw ‹ œ 0 Ê 2È2xw # 8È2 yw œ 0 Ê Parabola
yw ‹ 2È2 Š
È2 #
xw
È2 #
yw ‹ 8 Š
24. cot 2! œ
AC B
œ
00 1
8Š
1 2
œ 0 Ê 2! œ
È2 #
xw
œ 0 Ê 2! œ
y œ xw sin ! yw cos ! Ê x œ
È2 #
1 2
xw
; therefore x œ xw cos ! yw sin !,
È2 w È2 w # x # y È2 w È2 w È2 w # y ‹Š # x # y ‹
Ê !œ È2 #
1 4
1 4
È2 Š
È2 #
xw
È2 #
yw ‹
#
; therefore x œ xw cos ! yw sin !,
yw , y œ
È2 #
xw
È2 #
yw
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 10.3 Quadratic Equations and Rotations Ê Š
È2 #
È2 #
xw
yw ‹ Š
È2 #
xw
È2 #
yw ‹ Š
È2 #
xw
È2 #
yw ‹ Š
È2 #
xw
È2 #
639
yw ‹ 1 œ 0 Ê xw # yw # 2È2 xw 2
œ 0 Ê Hyperbola 25. cot 2! œ
AC B
œ
33 2
œ 0 Ê 2! œ
y œ xw sin ! yw cos ! Ê x œ Ê 3Š
È2 #
xw
È2 #
#
yw ‹ 2 Š
È2 #
È2 #
1 2
xw
xw
1 4
Ê !œ È2 #
È2 #
; therefore x œ xw cos ! yw sin !, È2 #
yw , y œ
yw ‹ Š
È2 #
xw +
xw È2 #
È2 #
yw
yw ‹ 3 Š
È2 #
xw
È2 #
#
yw ‹ œ 19 Ê 4xw # 2yw # œ 19
Ê Ellipse 26. cot 2! œ
AC B
œ
3 (1) 4È 3
œ
" È3
Ê 2! œ
È3 # È3 È 4 3Š #
y œ xw sin ! yw cos ! Ê x œ Ê 3Š
È3 #
#
xw 1# yw ‹
1 3
Ê !œ
xw #1 yw , y œ
1 #
1 6
; therefore x œ xw cos ! yw sin !,
xw
xw 1# yw ‹ Š 1# xw
È3 #
È3 #
yw
yw ‹ Š 1# xw
È3 #
#
yw ‹ œ 7 Ê 5xw # 3yw # œ 7
Ê Hyperbola 27. cot 2! œ
14 2 16
œ
3 4
Ê cos 2! œ
2! and cos ! œ É 1 cos œÉ #
28. cot 2! œ œÉ
AC B
1 ˆ 35 ‰ #
29. tan 2! œ
œ
œ
" 13 w
4" 4
2 È5
1ˆ 35 ‰ #
3 5
2! (if we choose 2! in Quadrant I); thus sin ! œ É 1 cos œÉ 2
œ
2 È5
(or sin ! œ
2 È5
and cos ! œ
1 ˆ 35 ‰ #
œ
" È5
" È5 )
2! œ 34 Ê cos 2! œ 35 (if we choose 2! in Quadrant II); thus sin ! œ É 1 cos 2
2! and cos ! œ É 1 cos œÉ #
1 ˆ 35 ‰ #
œ
1 È5
(or sin ! œ
1 È5
and cos ! œ
2 È5 )
" #
Ê 2! ¸ 26.57° Ê ! ¸ 13.28° Ê sin ! ¸ 0.23, cos ! ¸ 0.97; then Aw ¸ 0.9, Bw ¸ 0.0,
œ
" 5
œ
Cw ¸ 3.1, D ¸ 0.7, Ew ¸ 1.2, and Fw œ 3 Ê 0.9 xw # 3.1 yw # 0.7xw 1.2yw 3 œ 0, an ellipse 30. tan 2! œ
" 2 (3)
Ê 2! ¸ 11.31° Ê ! ¸ 5.65° Ê sin ! ¸ 0.10, cos ! ¸ 1.00; then Aw ¸ 2.1, Bw ¸ 0.0,
Cw ¸ 3.1, Dw ¸ 3.0, Ew ¸ 0.3, and Fw œ 7 Ê 2.1 xw # 3.1 yw # 3.0xw 0.3yw 7 œ 0, a hyperbola 31. tan 2! œ
4 14 w
œ
Ê 2! ¸ 53.13° Ê ! ¸ 26.5(° Ê sin ! ¸ 0.45, cos ! ¸ 0.89; then Aw ¸ 0.0, Bw ¸ 0.0,
4 3
Cw ¸ 5.0, D ¸ 0, Ew ¸ 0, and Fw œ 5 Ê 5.0 yw # 5 œ 0 or yw œ „ 1.0, parallel lines 32. tan 2! œ
12 2 18 w
œ
3 4
Ê 2! ¸ 36.87° Ê ! ¸ 18.43° Ê sin ! ¸ 0.32, cos ! ¸ 0.95; then Aw ¸ 0.0, Bw ¸ 0.0,
Cw ¸ 20.1, D ¸ 0, Ew ¸ 0, and Fw œ 49 Ê 20.1 yw # 49 œ 0, parallel lines 33. tan 2! œ
œ 5 Ê 2! ¸ 78.69° Ê ! ¸ 39.35° Ê sin ! ¸ 0.63, cos ! ¸ 0.77; then Aw ¸ 5.0, Bw ¸ 0.0,
5 3 2
Cw ¸ 0.05, Dw ¸ 5.0, Ew ¸ 6.2, and Fw œ 1 Ê 5.0 xw # 0.05 yw # 5.0xw 6.2yw 1 œ 0, a hyperbola 34. tan 2! œ
7 29 w
œ 1 Ê 2! ¸ 45.00° Ê ! ¸ 22.5° Ê sin ! ¸ 0.38, cos ! ¸ 0.92; then Aw ¸ 0.5, Bw ¸ 0.0,
Cw ¸ 10.4, D ¸ 18.4, Ew ¸ 7.6, and Fw œ 86 Ê 0.5 xw # 10.4ayw b# 18.4xw 7.6yw 86 œ 0, an ellipse 35. ! œ 90° Ê x œ xw cos 90° yw sin 90° œ yw and y œ xw sin 90° yw cos 90° œ xw (a)
xw # b#
yw # a#
œ1
(b)
yw # a#
xw # b#
œ1
(c) xw # yw # œ a#
(d) y œ mx Ê y mx œ 0 Ê D œ m and E œ 1; ! œ 90° Ê Dw œ 1 and Ew œ m Ê myw xw œ 0 Ê yw œ m" xw (e) y œ mx b Ê y mx b œ 0 Ê D œ m and E œ 1; ! œ 90° Ê Dw œ 1, Ew œ m and Fw œ b Ê myw xw b œ 0 Ê yw œ m" xw mb Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
640
Chapter 10 Conic Sections and Polar Coordinates
36. ! œ 180° Ê x œ xw cos 180° yw sin 180° œ xw and y œ xw sin 180° yw cos 180° œ yw (a)
xw # a#
yw # b#
œ1
(b)
xw # a#
yw # b#
(c) xw # yw # œ a#
œ1
(d) y œ mx Ê y mx œ 0 Ê D œ m and E œ 1; ! œ 180° Ê Dw œ m and Ew œ 1 Ê yw mxw œ 0 Ê yw œ mxw (e) y œ mx b Ê y mx b œ 0 Ê D œ m and E œ 1; ! œ 180° Ê Dw œ m, Ew œ 1 and Fw œ b Ê yw mxw b œ 0 Ê yw œ mxw b 37. (a) Aw œ cos 45° sin 45° œ Š " w# " # x # Aw œ "# , Cw œ
Ê
(b)
c a
œ
2È 2 #
œ
" #
, Bw œ 0, Cw œ cos 45° sin 45° œ "# , Fw œ 1
yw # œ 1 Ê xw # yw # œ 2 "# (see part (a) above), Dw œ Ew œ Bw œ 0, Fw œ a Ê
38. xy œ 2 Ê xw # yw # œ 4 Ê Ê eœ
È2 È2 # ‹Š # ‹
xw # 4
yw # 4
" #
xw # "# yw # œ a Ê xw # yw # œ 2a
œ 1 (see Exercise 37(b)) Ê a œ 2 and b œ 2 Ê c œ È4 4 œ 2È2
œ È2
39. Yes, the graph is a hyperbola: with AC 0 we have 4AC 0 and B# 4AC 0. 40. The one curve that meets all three of the stated criteria is the ellipse x# 4xy 5y# 1 œ 0. The reasoning: The symmetry about the origin means that (xß y) lies on the graph whenever (xß y) does. Adding Ax# Bxy Cy# Dx Ey F œ 0 and A(x)# B(x)(y) C(y)# D(x) E(y) F œ 0 and dividing the result by 2 produces the equivalent equation Ax# Bxy Cy# F œ 0. Substituting x œ 1, y œ 0 (because the point (1ß 0) lies on the curve) shows further that A œ F. Then Fx# Bxy Cy# F œ 0. By implicit differentiation, 2Fx By Bxyw 2Cyyw œ 0, so substituting x œ 2, y œ 1, and yw œ 0 (from Property 3) gives 4F B œ 0 Ê B œ 4F Ê the conic is Fx# 4Fxy Cy# F œ 0. Now substituting x œ 2 and y œ 1 again gives 4F 8F C F œ 0 Ê C œ 5F Ê the equation is now Fx# 4Fxy 5Fy# F œ 0. Finally, dividing through by F gives the equation x# 4xy 5y# 1 œ 0. 41. Let ! be any angle. Then Aw œ cos# ! sin# ! œ 1, Bw œ 0, Cw œ sin# ! cos# ! œ 1, Dw œ Ew œ 0 and Fw œ a# Ê xw # yw # œ a# . 42. If A œ C, then Bw œ B cos 2! (C A) sin 2! œ B cos 2!. Then ! œ
1 4
Ê 2! œ
1 #
Ê Bw œ B cos
1 #
œ 0 so the
xy-term is eliminated. 43. (a) B# 4AC œ 4# 4(1)(4) œ 0, so the discriminant indicates this conic is a parabola (b) The left-hand side of x# 4xy 4y# 6x 12y 9 œ 0 factors as a perfect square: (x 2y 3)# œ 0 Ê x 2y 3 œ 0 Ê 2y œ x 3; thus the curve is a degenerate parabola (i.e., a straight line). 44. (a) B# 4AC œ 6# 4(9)(1) œ 0, so the discriminant indicates this conic is a parabola (b) The left-hand side of 9x# 6xy y# 12x 4y 4 œ 0 factors as a perfect square: (3x y 2)# œ 0 Ê 3x y 2 œ 0 Ê y œ 3x 2; thus the curve is a degenerate parabola (i.e., a straight line).
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 10.3 Quadratic Equations and Rotations 45. (a) B# 4AC œ 1 4(0)(0) œ 1 Ê hyperbola (b) xy 2x y œ 0 Ê y(x 1) œ 2x Ê y œ (c) y œ
2x x1
Ê
dy dx
œ
2 (x 1)#
and we want
the slope of y œ 2x Ê 2 œ (x#1)
1 dy Š dx ‹
2x x1
œ 2,
#
Ê (x 1)# œ 4 Ê x œ 3 or x œ 1; x œ 3 Ê y œ 3 Ê (3ß 3) is a point on the hyperbola where the line with slope m œ 2 is normal Ê the line is y 3 œ 2(x 3) or y œ 2x 3; x œ 1 Ê y œ 1 Ê (1ß 1) is a point on the hyperbola where the line with slope m œ 2 is normal Ê the line is y 1 œ 2(x 1) or y œ 2x 3 46. (a) False: let A œ C œ 1, B œ 2 Ê B# 4AC œ 0 Ê parabola (b) False: see part (a) above (c) True: AC 0 Ê 4AC 0 Ê B# 4AC 0 Ê hyperbola 47. Assume the ellipse has been rotated to eliminate the xy-term Ê the new equation is Aw xw # Cw yw # œ 1 Ê the semi-axes are É A" and É C" Ê the area is 1 ŠÉ A" ‹ ŠÉ C" ‹ œ w
w
w
w
1 ÈA C
œ Bw # 4Aw Cw œ 4Aw Cw (because Bw œ 0) we find that the area is
w
w
œ
21 È4A C
21 È4AC B#
w
w
. Since B# 4AC
as claimed.
48. (a) Aw Cw œ aA cos# ! B cos ! sin ! C sin# !b aA sin# ! B cos ! sin ! C sin# !b œ A acos# ! sin# !b C asin# ! cos# !b œ A C (b) Dw # Ew # œ (D cos ! E sin !)# (D sin ! E cos !)# œ D# cos# ! 2DE cos ! sin ! E# sin# ! D# sin# ! 2DE sin ! cos ! E# cos# ! œ D# acos# ! sin# !b E# asin# ! cos# !b œ D# E# 49. Bw # 4Aw Cw œ aB cos 2! (C A) sin 2!b# 4 aA cos# ! B cos ! sin ! C sin# !b aA sin# ! B cos ! sin ! C cos# !b œ B# cos# 2! 2B(C A) sin 2! cos 2! (C A)# sin# 2! 4A# cos# ! sin# ! 4AB cos$ ! sin ! 4AC cos% ! 4AB cos ! sin$ ! 4B# cos# ! sin# ! 4BC cos$ ! sin ! 4AC sin% ! 4BC cos ! sin$ ! 4C# cos# ! sin# ! # œ B cos# 2! 2BC sin 2! cos 2! 2AB sin 2! cos 2! C# sin# 2! 2AC sin# 2! A# sin# 2! 4A# cos# ! sin# ! 4AB cos$ ! sin ! 4AC cos% ! 4AB cos ! sin$ ! B# sin# 2! 4BC cos$ ! sin ! 4AC sin% ! 4BC cos ! sin$ ! 4C# cos# ! sin# ! # œ B 2BC(2 sin ! cos !) acos# ! sin# !b 2AB(2 sin ! cos !) acos# ! sin# !b C# a4 sin# ! cos# !b 2AC a4 sin# ! cos# !b A# a4 sin# ! cos# !b 4A# cos# ! sin# ! 4AB cos$ ! sin ! 4AC cos% ! 4AB cos ! sin$ ! 4BC cos$ ! sin ! 4AC sin% ! 4BC cos ! sin$ ! 4C# cos# ! sin# ! # œ B 8AC sin# ! cos# ! 4AC cos% ! 4AC sin% ! œ B# 4AC acos% ! 2 sin# ! cos# ! sin% !b œ B# 4AC acos# ! sin# !b œ B# 4AC
#
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Chapter 10 Conic Sections and Polar Coordinates
10.4 CONICS AND PARAMETRIC EQUATIONS; THE CYCLOID 1. x œ cos t, y œ sin t, 0 Ÿ t Ÿ 1 Ê cos# t sin# t œ 1 Ê x# y# œ 1
2. x œ sin (21(1 t)), y œ cos (21(1 t)), 0 Ÿ t Ÿ 1 Ê sin# (21(1 t)) cos# (21(1 t)) œ 1 Ê x# y# œ 1
3. x œ 4 cos t, y œ 5 sin t, 0 Ÿ t Ÿ 1
4. x œ 4 sin t, y œ 5 cos t, 0 Ÿ t Ÿ 21
Ê
16 cos# t 16
25 sin# t 25
x# 16
œ1 Ê
y# 25
œ1
5. x œ t, y œ Èt, t 0 Ê y œ Èx
Ê
16 sin# t 16
25 cos# t 25
œ1 Ê
x# 16
6. x œ sec# t 1, y œ tan t, 1# t Ê sec# t 1 œ tan# t Ê x œ y#
7. x œ sec t, y œ tan t, 1# t #
#
#
1 # #
Ê sec t tan t œ 1 Ê x y œ 1
y# #5
œ1
1 #
8. x œ csc t, y œ cot t, 0 t 1 Ê 1 cot# t œ csc# t Ê 1 y# œ x# Ê x# y# œ 1
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Section 10.4 Conics and Parametric Equations; The Cycloid 9. x œ t, y œ È4 t# , 0 Ÿ t Ÿ 2 Ê y œ È4 x#
10. x œ t# , y œ Èt% 1, t 0 Ê y œ Èx# 1, x 0
11. x œ cosh t, y œ sinh t, _ 1 _ Ê cosh# t sinh# t œ 1 Ê x# y# œ 1
12. x œ 2 sinh t, y œ 2 cosh t, _ t _ Ê 4 cosh# t 4 sinh# t œ 4 Ê y# x# œ 4
13. Arc PF œ Arc AF since each is the distance rolled and Arc PF œ nFCP Ê Arc PF œ b(nFCP); ArcaAF œ ) b Ê Arc AF œ a) Ê a) œ b(nFCP) Ê nFCP œ nOCG œ
1 #
a b
);
); nOCG œ nOCP nPCE œ nOCP ˆ 1# !‰ . Now nOCP œ 1 nFCP œ 1 ba ). Thus nOCG œ 1 ba ) œ 1 ba )
1 #
1 #
! Ê
! Ê ! œ 1 ba ) ) œ 1
1 # ) ˆ ab b )‰ .
Then x œ OG BG œ OG PE œ (a b) cos ) b cos ! œ (a b) cos ) b cos ˆ1 œ (a b) cos ) b cos ˆ a b b )‰ . Also y œ EG œ CG CE œ (a b) sin ) b sin !
ab b
)‰
œ (a b) sin ) b sin ˆ1 a b b )‰ œ (a b) sin ) b sin ˆ a b b )‰ . Therefore x œ (a b) cos ) b cos ˆ a b b )‰ and y œ (a b) sin ) b sin ˆ a b b )‰ . If b œ 4a , then x œ ˆa 4a ‰ cos ) œ œ œ œ
3a 4 3a 4 3a 4 3a 4
cos )
œ œ œ
3a 4 3a 4 3a 4 3a 4
cos 3) œ
3a 4
cos Š
a ˆ 4a ‰ ˆ 4a ‰
)‹
cos ) 4a (cos ) cos 2) sin ) sin 2))
cos ) a(cos )) acos# ) sin# )b (sin ))(2 sin ) cos ))b a 2a # # 4 cos ) sin ) 4 sin ) cos ) # $ ) cos$ ) 3a 4 (cos )) a1 cos )b œ a cos ); a ˆ 4a ‰ a‰ a 3a a 3a 4 sin ) 4 sin Š ˆ 4a ‰ )‹ œ 4 sin ) 4 sin 3) œ 4
cos ) cos
y œ ˆa œ
a 4 a 4 a 4 a 4
a 4
cos$ )
sin ) 4a (sin ) cos 2) cos ) sin 2))
sin ) 4a a(sin )) acos# ) sin# )b (cos ))(2 sin ) cos ))b sin ) sin ) sin )
a 4 3a 4 3a 4
sin ) cos# ) sin ) cos# )
a 4 a 4 #
sin$ )
2a 4
cos# ) sin )
sin$ )
(sin )) a1 sin )b
a 4
sin$ ) œ a sin$ ).
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Chapter 10 Conic Sections and Polar Coordinates
14. P traces a hypocycloid where the larger radius is 2a and the smaller is a Ê x œ (2a a) cos ) a cos ˆ 2a a a )‰ œ 2a cos ), 0 Ÿ ) Ÿ 21, and y œ (2a a) sin ) a sin ˆ 2a a a )‰ œ a sin ) a sin ) œ 0. Therefore P traces the diameter of the circle back and forth as ) goes from 0 to 21. 15. Draw line AM in the figure and note that nAMO is a right angle since it is an inscribed angle which spans the diameter of a circle. Then AN# œ MN# AM# . Now, OA œ a, AN AM a œ tan t, and a œ sin t. Next MN œ OP Ê OP# œ AN# AM# œ a# tan# t a# sin# t Ê OP œ Èa# tan# t a# sin# t œ (a sin t)Èsec# t 1 œ a sin$ t cos t œ #
x œ OP sin t œ
a sin# t cos t #
. In triangle BPO,
a sin t tan t and
y œ OP cos t œ a sin t Ê x œ a sin# t tan t and y œ a sin# t. 16. Let the x-axis be the line the wheel rolls along with the y-axis through a low point of the trochoid (see the accompanying figure).
Let ) denote the angle through which the wheel turns. Then h œ a) and k œ a. Next introduce xw yw -axes parallel to the xy-axes and having their origin at the center C of the wheel. Then xw œ b cos ! and yw œ b sin !, where ! œ 3#1 ). It follows that xw œ b cos ˆ 3#1 )‰ œ b sin ) and yw œ b sin ˆ 3#1 )‰
œ b cos ) Ê x œ h xw œ a) b sin ) and y œ k yw œ a b cos ) are parametric equations of the trochoid.
# # # 17. D œ É(x 2)# ˆy "# ‰ Ê D# œ (x 2)# ˆy "# ‰ œ (t 2)# ˆt# "# ‰ Ê D# œ t% 4t
Ê
d aD # b dt
17 4
œ 4t$ 4 œ 0 Ê t œ 1. The second derivative is always positive for t Á 0 Ê t œ 1 gives a local
minimum for D# (and hence D) which is an absolute minimum since it is the only extremum Ê the closest point on the parabola is (1ß 1). # # 18. D œ Ɉ2 cos t 34 ‰ (sin t 0)# Ê D# œ ˆ2 cos t 34 ‰ sin# t Ê
d aD # b dt
œ 2 ˆ2 cos t 34 ‰ (2 sin t) 2 sin t cos t œ (2 sin t) ˆ3 cos t 3# ‰ œ 0 Ê 2 sin t œ 0 or 3 cos t Ê t œ 0, 1 or t œ #
#
1 3
,
51 3
. Now
#
#
d aD b dt#
œ 6 cos# t 3 cos t 6 sin# t so that #
#
#
d aD b dt#
3 #
œ0
(0) œ 3 Ê relative
#
maximum, d dtaD# b (1) œ 9 Ê relative maximum, d dtaD# b ˆ 13 ‰ œ 92 Ê relative minimum, and d # aD # b ˆ 5 1 ‰ œ 9# Ê relative minimum. Therefore both t œ 13 and t œ 531 give points on the ellipse dt# 3 È È the point ˆ 34 ß !‰ Ê Š1ß #3 ‹ and Š1ß #3 ‹ are the desired points.
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closest to
Section 10.4 Conics and Parametric Equations; The Cycloid 19. (a)
(b)
(c)
20. (a)
(b)
(c)
(b)
(c)
21.
22. (a)
23. (a)
(b)
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Chapter 10 Conic Sections and Polar Coordinates
24. (a)
25. (a)
(b)
(b)
(c)
26. (a)
(b)
(c)
(d)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 10.5 Polar Coordinates 10.5 POLAR COORDINATES 1. a, e; b, g; c, h; d, f
2. a, f; b, h; c, g; d, e
3. (a) ˆ2ß 1# 2n1‰ and ˆ2ß 1# (2n 1)1‰ , n an integer
(b) (#ß 2n1) and (#ß (2n 1)1), n an integer (c) ˆ2ß 3#1 2n1‰ and ˆ2ß 3#1 (2n 1)1‰ , n an integer
(d) (#ß (2n 1)1) and (#ß 2n1), n an integer
4. (a) ˆ3ß 14 2n1‰ and ˆ3ß 541 2n1‰ , n an integer (b) ˆ3ß 14 2n1‰ and ˆ3ß 541 2n1‰ , n an integer (c) ˆ3ß 14 2n1‰ and ˆ3ß 341 2n1‰ , n an integer (d) ˆ3ß 14 2n1‰ and ˆ3ß 341 2n1‰ , n an integer
5. (a) x œ r cos ) œ 3 cos 0 œ 3, y œ r sin ) œ 3 sin 0 œ 0 Ê Cartesian coordinates are ($ß 0) (b) x œ r cos ) œ 3 cos 0 œ 3, y œ r sin ) œ 3 sin 0 œ 0 Ê Cartesian coordinates are ($ß 0) (c) x œ r cos ) œ 2 cos 21 œ 1, y œ r sin ) œ 2 sin 21 œ È3 Ê Cartesian coordinates are Š1ß È3‹ 3
(d) x œ r cos ) œ 2 cos
71 3
3
œ 1, y œ r sin ) œ 2 sin
71 3
œ È3 Ê Cartesian coordinates are Š1ß È3‹
(e) x œ r cos ) œ 3 cos 1 œ 3, y œ r sin ) œ 3 sin 1 œ 0 Ê Cartesian coordinates are (3ß 0) (f) x œ r cos ) œ 2 cos 1 œ 1, y œ r sin ) œ 2 sin 1 œ È3 Ê Cartesian coordinates are Š1ß È3‹ 3
3
(g) x œ r cos ) œ 3 cos 21 œ 3, y œ r sin ) œ 3 sin 21 œ 0 Ê Cartesian coordinates are (3ß 0) (h) x œ r cos ) œ 2 cos ˆ 1 ‰ œ 1, y œ r sin ) œ 2 sin ˆ 1 ‰ œ È3 Ê Cartesian coordinates are Š1ß È3‹ 3
6. (a) x œ È2 cos
1 4
œ 1, y œ È2 sin
3
1 4
œ 1 Ê Cartesian coordinates are (1ß 1)
(b) x œ 1 cos 0 œ 1, y œ 1 sin 0 œ 0 Ê Cartesian coordinates are (1ß 0) (c) x œ 0 cos 1# œ 0, y œ 0 sin 1# œ 0 Ê Cartesian coordinates are (!ß 0) (d) x œ È2 cos ˆ 1 ‰ œ 1, y œ È2 sin ˆ 1 ‰ œ 1 Ê Cartesian coordinates are (1ß 1) 4
(e) x œ 3 cos
51 6
œ
4
3È 3 2
, y œ 3 sin
51 6
È
œ 3# Ê Cartesian coordinates are Š 3 # 3 ß 3# ‹
(f) x œ 5 cos ˆtan" 43 ‰ œ 3, y œ 5 sin ˆtan" 43 ‰ œ 4 Ê Cartesian coordinates are ($ß 4) (g) x œ 1 cos 71 œ 1, y œ 1 sin 71 œ 0 Ê Cartesian coordinates are (1ß 0) (h) x œ 2È3 cos 231 œ È3, y œ 2È3 sin 231 œ 3 Ê Cartesian coordinates are ŠÈ3ß 3‹
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7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 10.5 Polar Coordinates
649
22.
23. r cos ) œ 2 Ê x œ 2, vertical line through (#ß 0)
24. r sin ) œ 1 Ê y œ 1, horizontal line through (0ß 1)
25. r sin ) œ 0 Ê y œ 0, the x-axis
26. r cos ) œ 0 Ê x œ 0, the y-axis
27. r œ 4 csc ) Ê r œ
4 sin )
28. r œ 3 sec ) Ê r œ
Ê r sin ) œ 4 Ê y œ 4, a horizontal line through (0ß 4)
3 cos )
Ê r cos ) œ 3 Ê x œ 3, a vertical line through (3ß 0)
29. r cos ) r sin ) œ 1 Ê x y œ 1, line with slope m œ 1 and intercept b œ 1 30. r sin ) œ r cos ) Ê y œ x, line with slope m œ 1 and intercept b œ 0 31. r# œ 1 Ê x# y# œ 1, circle with center C œ (!ß 0) and radius 1 32. r# œ 4r sin ) Ê x# y# œ 4y Ê x# y# 4y 4 œ 4 Ê x# (y 2)# œ 4, circle with center C œ (0ß 2) and radius 2 33. r œ
5 sin )2 cos )
Ê r sin ) 2r cos ) œ 5 Ê y 2x œ 5, line with slope m œ 2 and intercept b œ 5
34. r# sin 2) œ 2 Ê 2r# sin ) cos ) œ 2 Ê (r sin ))(r cos )) œ 1 Ê xy œ 1, hyperbola with focal axis y œ x )‰ˆ " ‰ 35. r œ cot ) csc ) œ ˆ cos Ê r sin# ) œ cos ) Ê r# sin# ) œ r cos ) Ê y# œ x, parabola with vertex (0ß 0) sin ) sin )
which opens to the right sin ) ‰ 36. r œ 4 tan ) sec ) Ê r œ 4 ˆ cos Ê r cos# ) œ 4 sin ) Ê r# cos# ) œ 4r sin ) Ê x# œ 4y, parabola with #)
vertex œ (!ß 0) which opens upward
37. r œ (csc )) er cos ) Ê r sin ) œ er cos ) Ê y œ ex , graph of the natural exponential function 38. r sin ) œ ln r ln cos ) œ ln (r cos )) Ê y œ ln x, graph of the natural logarithm function 39. r# 2r# cos ) sin ) œ 1 Ê x# y# 2xy œ 1 Ê x# 2xy y# œ 1 Ê (x y)# œ 1 Ê x y œ „ 1, two parallel straight lines of slope 1 and y-intercepts b œ „ 1 40. cos# ) œ sin# ) Ê r# cos# ) œ r# sin# ) Ê x# œ y# Ê kxk œ kyk Ê „ x œ y, two perpendicular lines through the origin with slopes 1 and 1, respectively. 41. r# œ 4r cos ) Ê x# y# œ 4x Ê x# 4x y# œ 0 Ê x# 4x 4 y# œ 4 Ê (x 2)# y# œ 4, a circle with center C(2ß 0) and radius 2
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Chapter 10 Conic Sections and Polar Coordinates
42. r# œ 6r sin ) Ê x# y# œ 6y Ê x# y# 6y œ 0 Ê x# y# 6y 9 œ 9 Ê x# (y 3)# œ 9, a circle with center C(0ß 3) and radius 3 43. r œ 8 sin ) Ê r# œ 8r sin ) Ê x# y# œ 8y Ê x# y# 8y œ 0 Ê x# y# 8y 16 œ 16 Ê x# (y 4)# œ 16, a circle with center C(0ß 4) and radius 4 44. r œ 3 cos ) Ê r# œ 3r cos ) Ê x# y# œ 3x Ê x# y# 3x œ 0 Ê x# 3x # Ê ˆx 3# ‰ y# œ
9 4
, a circle with center C ˆ 3# ß !‰ and radius
9 4
y# œ
9 4
3 #
45. r œ 2 cos ) 2 sin ) Ê r# œ 2r cos ) 2r sin ) Ê x# y# œ 2x 2y Ê x# 2x y# 2y œ 0 Ê (x 1)# (y 1)# œ 2, a circle with center C(1ß 1) and radius È2 46. r œ 2 cos ) sin ) Ê r# œ 2r cos ) r sin ) Ê x# y# œ 2x y Ê x# 2x y# y œ 0 # Ê (x 1)# ˆy "# ‰ œ 54 , a circle with center C ˆ1ß "# ‰ and radius
È5 #
È
47. r sin ˆ) 16 ‰ œ 2 Ê r ˆsin ) cos 16 cos ) sin 16 ‰ œ 2 Ê #3 r sin ) "# r cos ) œ 2 Ê Ê È3 y x œ 4, line with slope m œ " and intercept b œ 4 È3
È3 #
È3
È
48. r sin ˆ 231 )‰ œ 5 Ê r ˆsin 231 cos ) cos 231 sin )‰ œ 5 Ê #3 r cos ) "# r sin ) œ 5 Ê Ê È3 x y œ 10, line with slope m œ È3 and intercept b œ 10 49. x œ 7 Ê r cos ) œ 7 51. x œ y Ê r cos ) œ r sin ) Ê ) œ
y "# x œ 2
È3 #
x "# y œ 5
50. y œ 1 Ê r sin ) œ 1 1 4
52. x y œ 3 Ê r cos ) r sin ) œ 3
53. x# y# œ 4 Ê r# œ 4 Ê r œ 2 or r œ 2 54. x# y# œ 1 Ê r# cos# ) r# sin# ) œ 1 Ê r# acos# ) sin# )b œ 1 Ê r# cos 2) œ 1 55.
x# 9
y# 4
œ 1 Ê 4x# 9y# œ 36 Ê 4r# cos# ) 9r# sin# ) œ 36
56. xy œ 2 Ê (r cos ))(r sin )) œ 2 Ê r# cos ) sin ) œ 2 Ê 2r# cos ) sin ) œ 4 Ê r# sin 2) œ 4 57. y# œ 4x Ê r# sin# ) œ 4r cos ) Ê r sin# ) œ 4 cos ) 58. x# xy y# œ 1 Ê x# y# xy œ 1 Ê r# r# sin ) cos ) œ 1 Ê r# (1 sin ) cos )) œ 1 59. x# (y 2)# œ 4 Ê x# y# 4y 4 œ 4 Ê x# y# œ 4y Ê r# œ 4r sin ) Ê r œ 4 sin ) 60. (x 5)# y# œ 25 Ê x# 10x 25 y# œ 25 Ê x# y# œ 10x Ê r# œ 10r cos ) Ê r œ 10 cos ) 61. (x 3)# (y 1)# œ 4 Ê x# 6x 9 y# 2y 1 œ 4 Ê x# y# œ 6x 2y 6 Ê r# œ 6r cos ) 2r sin ) 6 62. (x 2)# (y 5)# œ 16 Ê x# 4x 4 y# 10y 25 œ 16 Ê x# y# œ 4x 10y 13 Ê r# œ 4r cos ) 10r sin ) 13 63. (!ß )) where ) is any angle Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 10.6 Graphing in Polar Coordinates 64. (a) x œ a Ê r cos ) œ a Ê r œ (b) y œ b Ê r sin ) œ b Ê r œ
a cos ) b sin )
Ê r œ a sec ) Ê r œ b csc )
10.6 GRAPHING IN POLAR COORDINATES 1. 1 cos ()) œ 1 cos ) œ r Ê symmetric about the x-axis; 1 cos ()) Á r and 1 cos (1 )) œ 1 cos ) Á r Ê not symmetric about the y-axis; therefore not symmetric about the origin
2. 2 2 cos ()) œ 2 2 cos ) œ r Ê symmetric about the x-axis; 2 # cos ()) Á r and 2 2 cos (1 )) œ 2 2 cos ) Á r Ê not symmetric about the y-axis; therefore not symmetric about the origin
3. 1 sin ()) œ 1 sin ) Á r and 1 sin (1 )) œ 1 sin ) Á r Ê not symmetric about the x-axis; 1 sin (1 )) œ 1 sin ) œ r Ê symmetric about the y-axis; therefore not symmetric about the origin
4. 1 sin ()) œ 1 sin ) Á r and 1 sin (1 )) œ 1 sin ) Á r Ê not symmetric about the x-axis; 1 sin (1 )) œ 1 sin ) œ r Ê symmetric about the y-axis; therefore not symmetric about the origin
5. 2 sin ()) œ 2 sin ) Á r and 2 sin (1 )) œ 2 sin ) Á r Ê not symmetric about the x-axis; 2 sin (1 )) œ 2 sin ) œ r Ê symmetric about the y-axis; therefore not symmetric about the origin
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Chapter 10 Conic Sections and Polar Coordinates
6. 1 2 sin ()) œ 1 2 sin ) Á r and 1 2 sin (1 )) œ 1 2 sin ) Á r Ê not symmetric about the x-axis; 1 2 sin (1 )) œ 1 2 sin ) œ r Ê symmetric about the y-axis; therefore not symmetric about the origin
7. sin ˆ #) ‰ œ sin ˆ #) ‰ œ r Ê symmetric about the y-axis; sin ˆ 21#) ‰ œ sin ˆ 2) ‰ , so the graph is symmetric about the x-axis, and hence the origin.
8. cos ˆ #) ‰ œ cos ˆ #) ‰ œ r Ê symmetric about the x-axis; cos ˆ 21#) ‰ œ cos ˆ 2) ‰ , so the graph is symmetric about the y-axis, and hence the origin.
9. cos ()) œ cos ) œ r# Ê (rß )) and (rß )) are on the graph when (rß )) is on the graph Ê symmetric about the x-axis and the y-axis; therefore symmetric about the origin
10. sin (1 )) œ sin ) œ r# Ê (rß 1 )) and (rß 1 )) are on the graph when (rß )) is on the graph Ê symmetric about the y-axis and the x-axis; therefore symmetric about the origin
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Section 10.6 Graphing in Polar Coordinates 11. sin (1 )) œ sin ) œ r# Ê (rß 1 )) and (rß 1 )) are on the graph when (rß )) is on the graph Ê symmetric about the y-axis and the x-axis; therefore symmetric about the origin
12. cos ()) œ cos ) œ r# Ê (rß )) and (rß )) are on the graph when (rß )) is on the graph Ê symmetric about the x-axis and the y-axis; therefore symmetric about the origin
13. Since a „ rß )b are on the graph when (rß )) is on the graph ˆa „ rb# œ 4 cos 2( )) Ê r# œ 4 cos 2)‰ , the graph is symmetric about the x-axis and the y-axis Ê the graph is symmetric about the origin
14. Since (rß )) on the graph Ê (rß )) is on the graph ˆa „ rb# œ 4 sin 2) Ê r# œ 4 sin 2)‰ , the graph is symmetric about the origin. But 4 sin 2()) œ 4 sin 2) Á r# and 4 sin 2(1 )) œ 4 sin (21 2)) œ 4 sin (2)) œ 4 sin 2) Á r# Ê the graph is not symmetric about the x-axis; therefore the graph is not symmetric about the y-axis 15. Since (rß )) on the graph Ê (rß )) is on the graph ˆa „ rb# œ sin 2) Ê r# œ sin 2)‰ , the graph is symmetric about the origin. But sin 2()) œ ( sin 2)) sin 2) Á r# and sin 2(1 )) œ sin (21 2)) œ sin (2)) œ ( sin 2)) œ sin 2) Á r# Ê the graph is not symmetric about the x-axis; therefore the graph is not symmetric about the y-axis 16. Sincea „ rß )b are on the graph when (rß )) is on the graph ˆa „ rb# œ cos 2()) Ê r# œ cos 2)‰, the graph is symmetric about the x-axis and the y-axis Ê the graph is symmetric about the origin.
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Chapter 10 Conic Sections and Polar Coordinates Ê r œ 1 Ê ˆ1ß 1# ‰ , and ) œ 1# Ê r œ 1 w )r cos ) Ê ˆ1ß 1# ‰ ; rw œ ddr) œ sin ); Slope œ rrw sin cos )r sin )
17. ) œ
1 #
sin# )r cos ) sin ) cos )r sin ) sin# ˆ 1# ‰(1) cos 1# sin 1# cos 1# (1) sin 1#
œ
Ê Slope at ˆ1ß 1# ‰ is
œ 1; Slope at ˆ1ß 1# ‰ is
sin# ˆ 1# ‰(1) cos ˆ 1# ‰ sin ˆ 1# ‰ cos ˆ 1# ‰(1) sin ˆ 1# ‰
œ1
18. ) œ 0 Ê r œ 1 Ê ("ß 0), and ) œ 1 Ê r œ 1 dr Ê ("ß 1); rw œ d) œ cos );
rw sin )r cos ) cos ) sin )r cos ) rw cos )r sin ) œ cos ) cos )r sin ) 0 sin 0(1) cos 0 cos ) sin )r cos ) Ê Slope at ("ß 0) is coscos # 0(1) sin 0 cos# )r sin ) cos 1 sin 1(1) cos 1 1; Slope at ("ß 1) is cos# 1(1) sin 1 œ 1
Slope œ œ œ
Ê r œ 1 Ê ˆ"ß 14 ‰ ; ) œ 14 Ê r œ 1 Ê ˆ1ß 14 ‰ ; ) œ 341 Ê r œ 1 Ê ˆ"ß 341 ‰ ; ) œ 341 Ê r œ 1 Ê ˆ1ß 341 ‰ ;
19. ) œ
rw œ
1 4
dr d)
œ 2 cos 2);
Slope œ
r sin )r cos ) r cos )r sin ) w w
Ê Slope at ˆ1ß 14 ‰ is Slope at ˆ1ß 14 ‰ is Slope at ˆ1ß 341 ‰ is Slope at ˆ1ß 341 ‰ is
2 cos 2) sin )r cos ) 2 cos 2) cos )r sin ) 2 cos ˆ 1# ‰ sin ˆ 14 ‰(1) cos ˆ 14 ‰ 2 cos ˆ 1 ‰ cos ˆ 1 ‰(1) sin ˆ 1 ‰
œ
#
4
4
œ 1;
2 cos ˆ 1# ‰ sin ˆ 14 ‰(1) cos ˆ 14 ‰ 2 cos ˆ 1# ‰ cos ˆ 14 ‰(1) sin ˆ 14 ‰
2 cos Š 3#1 ‹ sin Š 341 ‹(1) cos Š 341 ‹ 2 cos Š 3#1 ‹ cos Š 341 ‹(1) sin Š 341 ‹
œ 1;
œ 1;
2 cos Š 3#1 ‹ sin Š 341 ‹(1) cos Š 341 ‹ 2 cos Š 3#1 ‹ cos Š 341 ‹(1) sin Š 341 ‹
œ 1
20. ) œ 0 Ê r œ 1 Ê (1ß 0); ) œ 12 Ê r œ 1 Ê ˆ1ß 12 ‰ ; ) œ 1# Ê r œ 1 Ê ˆ"ß 12 ‰ ; ) œ 1 Ê r œ 1 Ê (1ß 1); rw œ
dr d) œ 2 sin 2); )r cos ) 2 sin 2) sin )r cos ) Slope œ rr sin cos )r sin ) œ 2 sin 2) cos )r sin ) 2 sin 0 sin 0cos 0 Ê Slope at (1ß 0) is 2 sin 0 cos 0sin 0 , which is undefined; 2 sin 2 ˆ 1 ‰ sin ˆ 1 ‰(1) cos ˆ 1 ‰ Slope at ˆ1ß 12 ‰ is 2 sin 2 ˆ 12 ‰ cos ˆ21 ‰(1) sin ˆ 21 ‰ œ 0; w w
2
Slope at ˆ1ß 12 ‰ is Slope at ("ß 1) is
2
2
2 sin 2 ˆ 1# ‰ sin ˆ 1# ‰(1) cos ˆ 1# ‰ 2 sin 2 ˆ 1 ‰ cos ˆ 1 ‰(1) sin ˆ 1 ‰ #
2 sin 21 sin 1cos 1 2 sin 21 cos 1sin 1
#
#
œ 0;
, which is undefined
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Section 10.6 Graphing in Polar Coordinates 21. (a)
(b)
22. (a)
(b)
23. (a)
(b)
24. (a)
(b)
25.
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Chapter 10 Conic Sections and Polar Coordinates
26. r œ 2 sec ) Ê r œ
2 cos )
Ê r cos ) œ 2 Ê x œ 2
27.
28.
29. ˆ#ß 341 ‰ is the same point as ˆ2ß 14 ‰ ; r œ 2 sin 2 ˆ 14 ‰ œ 2 sin ˆ 1# ‰ œ 2 Ê ˆ2ß 14 ‰ is on the graph Ê ˆ#ß 341 ‰ is on the graph 30. ˆ "# ß 321 ‰ is the same point as ˆ "# ß 12 ‰ ; r œ sin Š
ˆ 1# ‰ 3 ‹
œ sin
1 6
œ "# Ê ˆ "# ß 1# ‰ is on the graph Ê ˆ "# ß 3#1 ‰
is on the graph 31. 1 cos ) œ 1 cos ) Ê cos ) œ 0 Ê ) œ 1# , 3#1 Ê r œ 1; points of intersection are ˆ"ß 1# ‰ and ˆ"ß 3#1 ‰ . The point of intersection (!ß 0) is found by graphing.
32. 1 sin ) œ 1 sin ) Ê sin ) œ 0 Ê ) œ 0, 1 Ê r œ 1; points of intersection are (1ß 0) and (1ß 1). The point of intersection (!ß 0) is found by graphing.
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Section 10.6 Graphing in Polar Coordinates 33. 2 sin ) œ 2 sin 2) Ê sin ) œ sin 2) Ê sin ) œ 2 sin ) cos ) Ê sin ) 2 sin ) cos ) œ 0 Ê (sin ))(1 2 cos )) œ 0 Ê sin ) œ 0 or cos ) œ
, or 13 ; ) œ 0 or 1 Ê r œ 0, Ê r œ È3 , and ) œ 1 Ê r œ È3 ; points of
Ê ) œ 0, 1, )œ
1 3
1 3
" #
3
intersection are (!ß 0), ŠÈ3 ß 13 ‹, and ŠÈ3 ß 13 ‹
34. cos ) œ 1 cos ) Ê 2 cos ) œ 1 Ê cos ) œ
" #
Ê ) œ 13 , 13 Ê r œ "# ; points of intersection are ˆ "# ß 13 ‰ and ˆ "# , 13 ‰ . The point (0ß 0) is found by graphing.
#
35. ŠÈ2‹ œ 4 sin ) Ê
" #
œ sin ) Ê ) œ
1 6
,
51 6
; points
of intersection are ŠÈ2ß 16 ‹ and ŠÈ2ß 561 ‹ . The points ŠÈ2ß 16 ‹ and ŠÈ2ß 561 ‹ are found by graphing.
36. È2 sin ) œ È2 cos ) Ê sin ) œ cos ) Ê ) œ )œ
1 4
#
Ê r œ 1 Ê r œ „ 1 and ) œ
51 4
1 4
,
51 4
;
#
Ê r œ 1
Ê no solution for r; points of intersection are ˆ „ 1ß 14 ‰ . The points (!ß 0) and ˆ „ 1ß 341 ‰ are found by graphing.
37. 1 œ 2 sin 2) Ê sin 2) œ
" #
Ê 2) œ
1 6
,
51 6
,
131 6
,
171 6
1 Ê ) œ 12 , 5121 , 13121 , 17121 ; points of intersection are 1‰ ˆ ˆ1ß 12 , 1ß 5121 ‰ , ˆ1ß 131#1 ‰, and ˆ1ß 171#1 ‰ . No other
points are found by graphing.
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Chapter 10 Conic Sections and Polar Coordinates
38. È2 cos 2) œ È2 sin 2) Ê cos 2) œ sin 2) Ê 2) œ 14 , 541 , 941 , 1341 Ê ) œ 18 , 581 , 981 , )œ
1 91 8 , 8 #
#
Ê r œ 1 Ê r œ „ 1; ) œ
51 8
,
131 8 131 8
;
Ê r œ 1 Ê no solution for r; points of intersection are ˆ1ß 18 ‰ and ˆ1ß 981 ‰ . The point of intersection (!ß 0) is found by graphing.
39. r# œ sin 2) and r# œ cos 2) are generated completely for 0 Ÿ ) Ÿ 1# . Then sin 2) œ cos 2) Ê 2) œ 14 is the only solution on that interval Ê ) œ 18 Ê r# œ sin 2 ˆ 18 ‰ œ È" Ê rœ „
" % È 2
2
" ß 1‹. % È 2 8
; points of intersection are Š „
The point of intersection (!ß 0) is found by graphing.
40. 1 sin
) #
31 #
Ê )œ )œ
71 #
œ 1 cos ,
71 #
) #
;)œ
Ê sin 31 #
Ê r œ 1 cos
intersection are Š"
) #
œ cos
Ê r œ 1 cos 71 4
œ1
È 2 31 # ß # ‹
È2 #
) #
Ê
) #
31 4
œ1
31 71 4 , 4 È2 # ;
œ
; points of
and Š1
È 2 71 # ß # ‹.
three points of intersection (0ß 0) and Š1 „
È2 #
The
ß 1# ‹ are
found by graphing and symmetry.
41. 1 œ 2 sin 2) Ê sin 2) œ
" #
Ê 2) œ
1 6
,
51 6
,
131 6
,
171 6
Ê ) œ 11# , 511# , 131#1 , 171#1 ; points of intersection are ˆ"ß 11# ‰ , ˆ"ß 511# ‰ , ˆ1ß 131#1 ‰ , and ˆ"ß 17121 ‰ . The points of intersection ˆ1ß 711# ‰ , ˆ"ß 111#1 ‰ , ˆ"ß 191#1 ‰ and ˆ"ß 231#1 ‰ are found by graphing and symmetry.
42. r# œ 2 sin 2) is completely generated on 0 Ÿ ) Ÿ " #
1 6
1 #
so
1 that 1 œ 2 sin 2) Ê sin 2) œ Ê 2) œ , 561 Ê ) œ 12 51 1 5 1 ˆ ‰ ˆ ‰ 1# ; points of intersection are 1ß 1# and "ß 1# . The 1 5 1 points of intersection ˆ"ß 1# ‰ and ˆ1ß 1# ‰ are found
,
by graphing.
43. Note that (rß )) and (rß ) 1) describe the same point in the plane. Then r œ 1 cos ) Í 1 cos () 1) œ 1 (cos ) cos 1 sin ) sin 1) œ 1 cos ) œ (1 cos )) œ r; therefore (rß )) is on the graph of r œ 1 cos ) Í (rß ) 1) is on the graph of r œ 1 cos ) Ê the answer is (a).
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Section 10.6 Graphing in Polar Coordinates
44. Note that (rß )) and (rß ) 1) describe the same point in the plane. Then r œ cos 2) Í sin ˆ2() 1)) 1# ‰ œ sin ˆ2) 5#1 ‰ œ sin (2)) cos ˆ 5#1 ‰ cos (2)) sin ˆ 5#1 ‰ œ cos 2) œ r; therefore (rß )) is on the graph of r œ sin ˆ2) 1# ‰ Ê the answer is (a).
45.
47. (a)
46.
(b)
(c)
(d)
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Chapter 10 Conic Sections and Polar Coordinates
48. (a)
(b)
(d)
(c)
(e)
#
#
49. (a) r# œ 4 cos ) Ê cos ) œ r4 ; r œ 1 cos ) Ê r œ 1 Š r4 ‹ Ê 0 œ r# 4r 4 Ê (r 2)# œ 0 #
Ê r œ 2; therefore cos ) œ 24 œ 1 Ê ) œ 1 Ê (2ß 1) is a point of intersection (b) r œ 0 Ê 0# œ 4 cos ) Ê cos ) œ 0 Ê ) œ 1# , 3#1 Ê ˆ!ß 1# ‰ or ˆ!ß 3#1 ‰ is on the graph; r œ 0 Ê 0 œ 1 cos ) Ê cos ) œ 1 Ê ) œ 0 Ê (0ß 0) is on the graph. Since (!ß 0) œ ˆ!ß 1# ‰ for polar coordinates, the graphs intersect at the origin. 50. (a) Let r œ f()) be symmetric about the x-axis and the y-axis. Then (rß )) on the graph Ê (rß )) is on the graph because of symmetry about the x-axis. Then (rß ())) œ (rß )) is on the graph because of symmetry about the y-axis. Therefore r œ f()) is symmetric about the origin. (b) Let r œ f()) be symmetric about the x-axis and the origin. Then (rß )) on the graph Ê (rß )) is on the graph because of symmetry about the x-axis. Then (rß )) is on the graph because of symmetry about the origin. Therefore r œ f()) is symmetric about the y-axis. (c) Let r œ f()) be symmetric about the y-axis and the origin. Then (rß )) on the graph Ê (rß )) is on the graph because of symmetry about the y-axis. Then ((r)ß )) œ (rß )) is on the graph because of symmetry about the origin. Therefore r œ f()) is symmetric about the x-axis. 51. The maximum width of the petal of the rose which lies along the x-axis is twice the largest y value of the curve on the interval 0 Ÿ ) Ÿ 14 . So we wish to maximize 2y œ 2r sin ) œ 2 cos 2) sin ) on 0 Ÿ ) Ÿ 14 . Let f()) œ 2 cos 2) sin ) œ 2 a1 2 sin# )b (sin )) œ 2 sin ) 4 sin$ ) Ê f w ()) œ 2 cos ) 12 sin# ) cos ). Then f w ()) œ 0 Ê 2 cos ) 12 sin# ) cos ) œ 0 Ê (cos )) a1 6 sin# )b œ 0 Ê cos ) œ 0 or 1 6 sin# ) œ 0 Ê ) œ sin ) œ
„1 È6 .
Since we want 0 Ÿ ) Ÿ
œ 2 Š È"6 ‹ 4 † interval 0 Ÿ ) Ÿ is
2È 6 9
" œ . We 6È 6 1 4 . Therefore the
1 4
"
, we choose ) œ sin
Š È"6 ‹
$
Ê f()) œ 2 sin ) 4 sin )
can see from the graph of r œ cos 2) that a maximum does occur in the maximum width occurs at ) œ sin" Š È"6 ‹ , and the maximum width
2È 6 9 .
52. We wish to maximize y œ r sin ) œ 2(1 cos ))(sin )) œ 2 sin ) 2 sin ) cos ). Then dy # # # d) œ 2 cos ) 2(sin ))( sin )) 2 cos ) cos ) œ 2 cos ) 2 sin ) 2 cos ) œ 2 cos ) 4 cos ) 2; thus dy # # d) œ 0 Ê 4 cos ) 2 cos ) 2 = 0 Ê 2 cos ) cos ) 1 œ 0 Ê (2 cos ) 1)(cos ) 1) œ 0 Ê cos ) œ or cos ) œ 1 Ê ) œ 13 , 531 , 1. From the graph, we can see that the maximum occurs in the first quadrant so È
we choose ) œ 13 . Then y œ 2 sin 13 2 sin 13 cos 13 œ 3 # 3 . The x-coordinate of this point is x œ r cos È œ 2 ˆ1 cos 13 ‰ ˆcos 13 ‰ œ 3# . Thus the maximum height is h œ 3 # 3 occurring at x œ 3# . Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
1 3
" #
1 #
or
Section 10.7 Area and Lengths in Polar Coordinates 10.7 AREA AND LENGTHS IN POLAR COORDINATES 1. A œ '0
21
" #
(4 2 cos ))# d) œ '0
21
" #
2) ‰‘ a16 16 cos ) 4 cos# )b d) œ '0 8 8 cos ) 2 ˆ 1 cos d) # 21
œ '0 (9 8 cos ) cos 2)) d) œ 9) 8 sin ) 21
2. A œ '0
21
œ
" #
a#
" #
[a(1 cos ))]# d) œ '0
21
" #
#1
" 2
sin 2)‘ ! œ 181
a# a1 2 cos ) cos# )b d) œ
" #
a#
2) ‰ '021 ˆ1 2 cos ) 1 cos d) #
'021 ˆ 3# 2 cos ) "# cos 2)‰ d) œ "# a# 3# ) 2 sin ) 4" sin 2)‘ #!1 œ 3# 1a#
3. A œ 2 '0
1Î4
" #
cos# 2) d) œ '0
" #
a2a# cos 2)b d) œ 2a# '1Î4 cos 2) d) œ 2a# sin22) ‘ 1Î% œ 2a#
1Î4
4. A œ 2 '1Î4 5. A œ '0
1Î2
" #
1Î4
1 cos 4) #
d) œ
" #
)
sin 4) ‘ 1Î% 4 !
œ
1Î4
(4 sin 2)) d) œ '0
1Î2
6. A œ (6)(2)'0
1Î6
1 8 1Î%
1Î#
2 sin 2) d) œ c cos 2)d !
œ2
(2 sin 3)) d) œ 12 '0 sin 3) d) œ 12 cos3 3) ‘ ! 1Î6
" #
1Î'
œ4
7. r œ 2 cos ) and r œ 2 sin ) Ê 2 cos ) œ 2 sin ) Ê cos ) œ sin ) Ê ) œ 14 ; therefore A œ 2 '0
1Î4
œ '0
1Î4
(2 sin ))# d) œ '0
1Î4
" #
2) ‰ 4 ˆ 1 cos d) œ '0 #
1Î4
œ c2) sin
1Î% 2) d !
œ
1 #
4 sin# ) d)
(2 2 cos 2)) d)
1
8. r œ 1 and r œ 2 sin ) Ê 2 sin ) œ 1 Ê sin ) œ Ê )œ
1 6
or
51 6
" #
; therefore
A œ 1(1)# '1Î6
51Î6
" #
c(2 sin ))# 1# d d)
œ 1 '1Î6 ˆ2 sin# ) "# ‰ d) 51Î6
œ 1 '1Î6 ˆ1 cos 2) "# ‰ d) 51Î6
œ 1 '1Î6 ˆ "# cos 2)‰ d) œ 1 "2 )
sin 2) ‘ &1Î' # 1Î'
œ 1 ˆ 511#
41 3È 3 6
51Î6
" #
sin
51 ‰ 3
1 ˆ 12
" #
sin 13 ‰ œ
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662
Chapter 10 Conic Sections and Polar Coordinates
9. r œ 2 and r œ 2(1 cos )) Ê 2 œ 2(1 cos )) Ê cos ) œ 0 Ê ) œ „ 1# ; therefore A œ 2 '0
1Î2
œ '0
1Î2
œ '0
1Î2
œ '0
1Î2
" #
[2(1 cos ))]# d) "# area of the circle
4 a1 2 cos ) cos# )b d) ˆ "# 1‰ (2)# 4 ˆ1 2 cos )
1 cos 2) ‰ #
d) 2 1
(4 8 cos ) 2 2 cos 2)) d) 21 1Î#
œ c6) 8 sin ) sin 2)d !
21 œ 51 8
10. r œ 2(1 cos )) and r œ 2(1 cos )) Ê 1 cos ) œ 1 cos ) Ê cos ) œ 0 Ê ) œ 1# or 3#1 ; the graph also gives the point of intersection (0ß 0); therefore A œ 2 '0
1Î2
" #
[2(1 cos ))]# d) 2 '1Î2 "# [2(1 cos ))]# d) 1
œ '0
4 a1 2 cos ) cos# )b d)
œ '0
4 ˆ1 2 cos )
œ '0
(6 8 cos ) 2 cos 2)) d) '1Î2 (6 8 cos ) 2 cos 2)) d)
1Î2
'1Î2 4 a1 2 cos ) cos# )b d) 1
1Î2 1Î2
1 cos 2) ‰ #
d) '1Î2 4 ˆ1 2 cos ) 1
1 cos 2) ‰ #
d)
1
1Î#
œ c6) 8 sin ) sin 2)d !
c6) 8 sin ) sin 2)d 11Î# œ 61 16
11. r œ È3 and r# œ 6 cos 2) Ê 3 œ 6 cos 2) Ê cos 2) œ 1 6
Ê )œ
" #
(in the 1st quadrant); we use symmetry of the
graph to find the area, so A œ 4 '0 ” "# (6 cos 2)) "# ŠÈ3‹ • d) #
1Î6
œ 2 '0 (6 cos 2) 3) d) œ 2 c3 sin 2) 3)d ! 1Î6
1Î'
œ 3È3 1 12. r œ 3a cos ) and r œ a(1 cos )) Ê 3a cos ) œ a(1 cos )) Ê 3 cos ) œ 1 cos ) Ê cos ) œ "# Ê ) œ 13 or 13 ; the graph also gives the point of intersection (0ß 0); therefore A œ 2 '0
1Î3
" #
c(3a cos ))# a# (1 cos ))# d d)
œ '0 a9a# cos# ) a# 2a# cos ) a# cos# )b d) 1Î3
œ '0
1Î3
a8a# cos# ) 2a# cos ) a# b d)
œ '0 c4a# (1 cos 2)) 2a# cos ) a# d d) 1Î3
œ '0 a3a# 4a# cos 2) 2a# cos )b d) 1Î3
1Î$
œ c3a# ) 2a# sin 2) 2a# sin )d !
œ 1a# 2a# ˆ "# ‰ 2a# Š
È3 # ‹
œ a# Š1 1 È3‹
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Section 10.7 Area and Lengths in Polar Coordinates
663
13. r œ 1 and r œ 2 cos ) Ê 1 œ 2 cos ) Ê cos ) œ "# Ê )œ
A œ 2'
1
21 3
in quadrant II; therefore c(2 cos ))# 1# d d) œ '21Î3 a4 cos# ) 1b d) 1
" 21Î3 #
œ '21Î3 [2(1 cos 2)) 1] d) œ '21Î3 (1 2 cos 2)) d) 1
1
œ c) sin 2)d 1#1Î$ œ 14. (a) A œ 2 '0
21Î3
œ '0
21Î3
" #
1 3
È3 #
(2 cos ) 1)# d) œ '0
21Î3
a4 cos# ) 4 cos ) 1b d) œ '0
21Î3
#1Î$
(3 2 cos 2) 4 cos )) d) œ c3) sin 2) 4 sin )d ! 3È 3 # ‹
(b) A œ Š21
Š1
3È 3 # ‹
1 6
; therefore A œ '1Î6
51Î6
51 6
or
œ '1Î6 ˆ18 51Î6
9 #
È3 #
4È 3 #
œ 21
3È 3 #
œ 1 3È3 (from 14(a) above and Example 2 in the text) " #
15. r œ 6 and r œ 3 csc ) Ê 6 sin ) œ 3 Ê sin ) œ Ê )œ
œ 21
[2(1 cos 2)) 4 cos ) 1] d)
csc# )‰ d) œ 18)
" #
a6# 9 csc# )b d)
9 #
cot )‘ 1Î'
&1Î'
œ Š151 9# È3‹ Š31 9# È3‹ œ 121 9È3
16. r# œ 6 cos 2) and r œ Ê
sec ) Ê
3 # #
9 4 %
sec# ) œ 6 cos 2) Ê
œ 2 cos% ) cos ) Ê 2 cos ) cos# )
3 8
1 6
(in the first quadrant); thus A œ 2 '0
œ 3 sin 2)
9 4
tan )‘ !
1Î6
1Î'
17. (a) r œ tan ) and r œ Š
œ 3Š
È2 # ‹
È3 # ‹
9 4È 3
œ
3È 3 #
csc ) Ê tan ) œ Š
" ˆ # 6 cos 2) 3È 3 3È 3 4 œ 4
È2 # ‹
È2 # ‹
cos ) Ê 1 cos# ) œ Š
Ê cos# ) Š
È2 # ‹
cos ) 1 œ 0 Ê cos ) œ È2 or 1 4
œ acos# )b a2 cos# ) 1b
9 4
È3 #
(the second equation has no real
sec# )‰ d) œ '0 ˆ6 cos 2) 1Î6
9 4
sec# )‰ d)
csc )
Ê sin# ) œ Š
(use the quadratic formula) Ê ) œ
3 8
or cos# ) œ "4 Ê cos ) œ „
3 4
roots) Ê ) œ
È2 #
œ cos# ) cos 2) Ê
œ 0 Ê 16 cos% ) 8 cos# ) 3 œ 0
3 8
Ê a4 cos# ) 1ba4 cos# ) 3b œ 0 Ê cos# ) œ
9 24
È2 # ‹
cos )
(the solution
in the first quadrant); therefore the area of R" is A" œ '0
1Î4
AO œ Š
" #
È2 # ‹
tan# ) d) œ 1 #
csc
œ
È2 #
" #
and OB œ Š
Ê the area of R# is A# œ 2 ˆ "#
1 8
4" ‰ œ
3 #
1 4
'01Î4 asec# ) 1b d)
" #
Š
rœ (b)
lim
lim
œ
" 4
1 4
1Î%
ctan ) )d !
œ
" #
ˆtan
œ 1 Ê AB œ Ê1# Š
1 4
14 ‰ œ
È2 # ‹
#
œ
) Ä 1 Î2 c
r œ sec ) as ) Ä
" ‰ cos ) 1c #
1 8
;
; therefore the area of the region shaded in the text is
1 4
generates the arc OB of r œ tan ) but does not generate the segment AB of the line
tan ) œ _ and the line x œ 1 is r œ sec ) in polar coordinates; then sin ) ˆ cos )
" #
È2 #
csc ). Instead the interval generates the half-line from B to _ on the line r œ
) Ä 1 Î2
=
È2 È2 # ‹Š # ‹
csc
" #
. Note: The area must be found this way since no common interval generates the region. For
example, the interval 0 Ÿ ) Ÿ È2 #
È2 # ‹
œ
œ
lim
) Ä 1 Î2 c
ˆ sincos) ) 1 ‰ œ
lim
) Ä 1 Î2 c
lim
) Ä 1 Î2 c
È2 #
csc ).
(tan ) sec ))
) ‰ ˆ cos sin ) œ 0 Ê r œ tan ) approaches
Ê r œ sec ) (or x œ 1) is a vertical asymptote of r œ tan ). Similarly, r œ sec )
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
664
Chapter 10 Conic Sections and Polar Coordinates (or x œ 1) is a vertical asymptote of r œ tan ).
18. It is not because the circle is generated twice from ) œ 0 to 21. The area of the cardioid is A œ 2 '0
1
" #
2) (cos ) 1)# d) œ '0 acos# ) 2 cos ) 1b d) œ '0 ˆ 1 cos 2 cos ) 1‰ d) # 1
œ 32) 31 #
sin 2) 4 1 51 œ 4 4
1
2 sin )‘ ! œ
19. r œ )# , 0 Ÿ ) Ÿ È5 Ê
#
È5
È5
œ '0 k)k È)# 4 d) œ (since ) 0) '0
) œ È5 Ê u œ 9“ Ä '4
9
20. r œ
e) È2
,0Ÿ)Ÿ1 Ê
dr d)
" #
œ
1 4
. The area of the circle is A œ 1 ˆ "# ‰ œ
œ 2); therefore Length œ '0
dr d)
È5
31 #
1
Èu du œ
e) È2
" #
Ê the area requested is actually
Éa)# b# (2))# d) œ '
0
) È ) # 4 d ) ; u œ ) # 4 Ê
23 u$Î# ‘ * œ %
È5
" #
È ) % 4) # d)
du œ ) d); ) œ 0 Ê u œ 4,
19 3
; therefore Length œ '0 ÊŠ Èe 2 ‹ Š Èe 2 ‹ d) œ '0 Ê2 Š e# ‹ d) 1
#
)
#
)
1
2)
œ '0 e) d) œ e) ‘ ! œ e1 1 1
1
21. r œ 1 cos ) Ê
dr d)
œ sin ); therefore Length œ '0 È(1 cos ))# ( sin ))# d) 21
1 œ 2 '0 È2 2 cos ) d) œ 2'0 É 4(1 #cos )) d) œ 4 '0 É 1 #cos ) d) œ 4 '0 cos ˆ #) ‰ d) œ 4 2 sin 2) ‘ ! œ 8 1
1
22. r œ a sin#
) #
1
, 0 Ÿ ) Ÿ 1, a 0 Ê
œ '0 Éa# sin% 1
) #
a# sin#
) #
dr d) ) #
cos#
œ a sin
) #
cos
) #
1
# ; therefore Length œ '0 Ɉa sin# #) ‰ ˆa sin 1
d) œ '0 a ¸sin #) ¸ Ésin# 1
) #
) #
cos#
1
6 1 cos )
œ '0
1Î2
,0Ÿ)Ÿ
1 #
É (1 36 cos ))#
œ ˆsince
" 1 cos )
Ê
dr d)
œ
; therefore Length œ '0
1Î2
6 sin ) (1 cos ))#
d) œ 6 '0
1Î2
36 sin# ) a1 cos )b%
" ¸ 1cos ¸ ) É1
0
1Î2
1Î2
1Î2
1Î2
) #
d)
cos# ) sin# ) 0 on 0 Ÿ ) Ÿ 1# ‰ 6 '0 ˆ 1 "cos ) ‰ É 1 2 cos(1)cos d) ) )#
cos ) È ' œ 6 '0 ˆ 1 "cos ) ‰ É (12 2cos ) )# d) œ 6 2 0
œ 3'0 sec$
#
#
6 sin ) ʈ 1 6cos ) ‰ Š (1 cos ))# ‹ d)
sin# ) (1 cos ))#
d) œ 6'0
1Î4
d) (1 cos ))$Î#
œ 6È2 '0
1Î2
1Î% sec$ u du œ (use tables) 6 Œ sec u2tan u ‘ !
d) ˆ2 cos# #) ‰$Î# " #
'01Î4
œ 3'0
1Î2
¸sec$ #) ¸ d)
sec u du
1Î% œ 6 Š È"2 2" ln ksec u tan uk‘ ! ‹ œ 3 ’È2 ln Š1 È2‹“
24. r œ
2 1 cos )
,
1 #
Ÿ)Ÿ1 Ê
4 œ '1Î2 Ê (1 cos ) ) # Š1 1
dr d)
œ
2 sin ) (1 cos ))#
sin# ) ‹ a1 cos )b#
œ ˆsince 1 cos ) 0 on
1 #
sin ) ; therefore Length œ '1Î2 ʈ 1 2cos ) ‰ Š (12cos ))# ‹ d) 1
1
#
1
œ '1Î2 csc$ ˆ #) ‰ d) œ ˆsince csc 1
) #
cos ) d) È ' È ' œ 2 '1Î2 ˆ 1 "cos ) ‰ É (12 2cos ))# d) œ 2 2 1Î2 (1 cos ))$Î# œ 2 2 1Î2 1
#
d)
cos ) sin Ÿ ) Ÿ 1‰ 2 '1Î2 ˆ 1 "cos ) ‰ É 1 2 cos(1)cos ) )# 1
) #
0 on
1 #
1
#
#
) sin d) œ '1Î2 ¸ 1 2cos ) ¸ É (1 (1cos )cos ) )#
#
d) ˆ2 sin# )# ‰$Î#
)
d)
œ '1Î2 ¸csc$ #) ¸ d) 1
Ÿ ) Ÿ 1‰ 2 '1Î4 csc$ u du œ (use tables) 1Î2
# cos #) ‰ d)
d) œ (since 0 Ÿ ) Ÿ 1) a ' sin ˆ #) ‰ d)
1 œ 2a cos 2) ‘ ! œ 2a
23. r œ
) #
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 10.7 Area and Lengths in Polar Coordinates 1Î# 2Œ csc u2cot u ‘ 1Î%
'11ÎÎ42
" #
1Î#
csc u du œ 2 Š È"2 2" ln kcsc u cot uk‘ 1Î% ‹ œ 2 ’ È"2
" #
ln ŠÈ2 1‹“
œ È2 ln Š1 È2‹ ) 3
25. r œ cos$ œ '0
Ê
dr d)
) 3
œ sin
; therefore Length œ '0
1Î4
) 3
cos#
Écos' ˆ 3) ‰ sin# ˆ 3) ‰ cos% ˆ 3) ‰ d) œ '
1Î4
1Î4 1cos ˆ 2) ‰ 3
#
d) œ
" #
)
3 2
2) ‘ 1Î% 3 !
sin
26. r œ È1 sin 2) , 0 Ÿ ) Ÿ 1È2 Ê 1È 2
Length œ '0 œ '0
È
1 2
É(1 sin 2))
sin 2) ' É 212sin 2) d) œ 0
27. r œ È1 cos 2) Ê 1È 2
œ '0
È
1 2
œ
dr d)
" #
cos# 2) (1 sin 2))
È
d) œ '0
1 2
1È 2
d) œ '0
2)
1È# !
#
sin 2) cos É 1 2 sin 2)1 sin 2)
#
2)
d)
œ 21
È
1È 2
cos 2) ' É 212cos 2 ) d) œ 0
21
É(1 cos 2))
È2 d) œ ’È2 )“
œ 0; Length œ '0 Èa# 0# d) œ '0 kak d) œ ca)d #!1 œ 21a
1È# !
sin# 2) (1 cos 2))
21
œ a sin ); Length œ '0 È(a cos ))# (a sin ))# d) œ '0 Èa# acos# ) sin# )b d)
dr d)
œ a cos ); Length œ '0 È(a cos ))# (a sin ))# d) œ '0 Èa# acos# ) sin# )b d)
1
œ '0 kak d) œ ca)d 1! œ 1a 1
d)
œ 21
dr d)
(b) r œ a cos ) Ê
(c) r œ a sin ) Ê
cos# ˆ 3) ‰ d)
(1 sin 2))"Î# (2 cos 2)) œ (cos 2))(1 sin 2))"Î# ; therefore
È2 d) œ ’È2 )“
#
1Î4
3 8
1 2
#
dr d)
" #
œ
# cos# 3) ‰ d) 0
(1 cos 2))"Î# (2 sin 2)); therefore Length œ '0
cos 2) sin É 1 2 cos 21) cos 2)
28. (a) r œ a Ê
dr d)
1 8
œ
) 3
ˆcos# 3) ‰ Écos# ˆ 3) ‰ sin# ˆ 3) ‰ d) œ '
1Î4
0
œ '0
Ɉcos$ 3) ‰# ˆ sin
1
1
1
œ '0 kak d) œ ca)d 1! œ 1a 1
29. r œ Ècos 2) , 0 Ÿ ) Ÿ œ '0
1Î4
1 4
Ê
dr d)
œ
" #
(cos 2))"Î# ( sin 2))(2) œ
sin 2) Ècos 2)
; therefore Surface Area
sin 2) (21r cos )) ÊŠÈcos 2)‹ Š È ‹ d) œ '0 Š21Ècos 2)‹ (cos ))Écos 2) cos 2) #
#
œ '0 Š21Ècos 2)‹ (cos ))É cos" 2) d) œ '0 1Î4
1Î4
30. r œ È2e)Î2 , 0 Ÿ ) Ÿ
1 #
Ê
dr d)
œ È2 ˆ "# ‰ e)Î2 œ
œ '0 Š21È2 e)Î2 ‹ (sin )) ÊŠÈ2 e)Î2 ‹ Š #
1Î2
1Î4
1Î%
21 cos ) d) œ c21 sin )d ! È2 #
È2 #
sin# 2) cos 2)
d)
œ 1È2
e)Î2 ; therefore Surface Area
e)Î2 ‹ d) œ '0 Š21È2 e)Î2 ‹ (sin )) É2e) "# e) d) #
1Î2
œ '0 Š21È2 e)Î2 ‹ (sin )) É 5# e) d) œ '0 Š21È2 e)Î2 ‹ (sin )) Š È52 e)Î2 ‹ d) œ 21È5 '0 e) sin ) d) 1Î2
1Î2
)
1Î#
œ 21È5 e2 (sin ) cos ))‘ !
œ 1È5 ae1Î2 1b where we integrated by parts
31. r# œ cos 2) Ê r œ „ Ècos 2) ; use r œ Ècos 2) on 0ß 14 ‘ Ê therefore Surface Area œ 2 '0 Š21Ècos 2)‹ (sin )) Écos 2) 1Î4
œ 41 '0 sin ) d) œ 41 c cos )d ! 1Î4
1Î2
È
1Î%
œ 41 ’
È2 #
dr d)
œ
sin# 2) cos 2)
" #
(cos 2))"Î# ( sin 2))(2) œ
d) œ 41 '0
1Î4
sin 2) Ècos 2)
Ècos 2) (sin )) É
(1)“ œ 21 Š2 È2‹
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
;
" cos 2)
d)
665
666
Chapter 10 Conic Sections and Polar Coordinates
32. r œ 2a cos ) Ê
dr d)
œ 2a sin ); therefore Surface Area œ '0 21(2a cos ))(cos ))È(2a cos ))# (2a sin ))# d) 1
œ 4a1 '0 acos# )b È4a# acos# ) sin# )b d) œ 8a1 '0 acos# )b kak d) œ 8a# 1 '0 cos# ) d) 1
1
1
2) ‰ œ 8a# 1 '0 ˆ 1 cos d) œ 4a# 1 '0 (1 cos 2)) d) œ 4a# 1 ) # 1
1
33. Let r œ f()). Then x œ f()) cos ) Ê
" 2
1
sin 2)‘ ! œ 4a# 1#
‰# œ cf w ()) cos ) f()) sin )d# œ f w ()) cos ) f()) sin ) Ê ˆ dx d)
dx d)
œ cf w ())d# cos# ) 2f w ()) f()) sin ) cos ) [f())]# sin# ); y œ f()) sin ) Ê
dy d)
#
œ f w ()) sin ) f()) cos )
# # w w # w # # Ê Š dy d) ‹ œ cf ()) sin ) f()) cos )d œ cf ())d sin ) 2f ())f()) sin ) cos ) [f())] cos ). Therefore #
# # w # # # # # w # # ˆ dx ‰# Š dy ˆ dr ‰# d) d) ‹ œ cf ())d acos ) sin )b [f())] acos ) sin )b œ cf ())d [f())] œ r d) .
' Ér# ˆ ddr) ‰# d). ‰# Š dy Thus, L œ '! ʈ dx d) d) ‹ d) œ ! "
"
#
'021 a(1 cos )) d) œ 2a1 c) sin )d #!1 œ a 21 rav œ 21"0 '0 a d) œ #"1 ca)d #!1 œ a 1Î2 1Î# rav œ ˆ 1 ‰"ˆ 1 ‰ 'c1Î2 a cos ) d) œ 1" ca sin )d 1Î# œ 2a 1
34. (a) rav œ (b) (c)
" 2 1 0
#
#
35. r œ 2f()), ! Ÿ ) Ÿ " Ê
dr d)
œ 2f w ()) Ê r# ˆ ddr) ‰ œ [2f())]# c2f w ())d# Ê Length œ '! É4[f())]# 4 cf w ())d# d) "
#
œ 2 '! É[f())]# cf w ())d# d) which is twice the length of the curve r œ f()) for ! Ÿ ) Ÿ " . "
36. Again r œ 2f()) Ê r# ˆ ddr) ‰ œ [2f())]2 c2f w ())d# Ê Surface Area œ '! 21[2f()) sin )] É4[f())]# 4 cf w ())d# d) "
#
œ 4 '! 21[f()) sin )] É[f())]# cf w ())d# d) which is four times the area of the surface generated by revolving "
r œ f()) about the x-axis for ! Ÿ ) Ÿ " . '021 r$ cos ) d) 37. x œ œ '021 r# d) 2 3
œ
2 3
2 3
'021 [a(1 cos ))]$ (cos )) d) œ '021 [a(1 cos ))]# d)
2 3
a$
'021 a1 3 cos ) 3 cos# ) cos$ )b (cos )) d) 21 a# '0 a1 2 cos ) cos# )b d) #
2) 1 cos 2) # a '0 ’cos ) 3 ˆ 1 cos # ‰ 3 a1 sin )b (cos )) ˆ # ‰ “ d) 21
'0
21
2) 1 2 cos ) ˆ 1 cos # ‰‘ d )
œ (After considerable algebra using
" 4 # ' 15 8 1 cos 2A ‰ a 0 ˆ 12 3 cos ) 3 cos 2) 2 cos ) sin ) 12 cos 4)‰ d) 21 # " 3 ' ˆ # 2 cos ) # cos 2)‰ d) 21
the identity cos# A œ
0
œ
#1
" 8 2 2 $ ‘ a 15 12 ) 3 sin ) 3 sin 2) 3 sin ) 48 sin 4) ! #3 ) 2 sin ) "4 sin 2)‘ #1
yœ
!
2 3
2 3
'2a2a "a u$ du 31
œ
0 31
'01 r# d) œ '01 a# d) œ ca# )d !1 œ a# 1; x œ yœ
2 3
œ
5 6
a;
21 2' $ '021 r$ sin ) d) 3 0 [a(1 cos ))] (sin )) d) œ ; u œ a(1 cos )) Ê "a du œ sin ) d); ) œ 0 Ê u œ 2a; 21 31 '0 r # d )
) œ 21 Ê u œ 2ad Ä 38.
œ
‰ a ˆ 15 6 1 31
2' $ '0 r$ sin ) d) 3 0 a sin ) d) œ a# 1 '01 r# d) œ 1
1
2 3
œ 0. Therefore the centroid is aBß yb œ ˆ 56 aß 0‰ 2 3
'01 r$ cos ) d) œ '01 r# d)
a$ c cos )d 1! a# 1
œ
ˆ 43 ‰ a$ a# 1
œ
2 3
'01 a$ cos ) d)
4a 31 .
a# 1
œ
2 3
a$ c sin )d 1! a# 1
œ
0 a# 1
œ 0;
Therefore the centroid is axß yb œ ˆ0ß 34a1 ‰ .
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 10.8 Conic Sections in Polar Coordinates
667
10.8 CONIC SECTIONS IN POLAR COORDINATES 1. r cos ˆ) 16 ‰ œ 5 Ê r ˆcos ) cos œ 10 Ê y œ È3 x 10 2. r cos ˆ) Ê
È2 #
3. r cos ˆ) Ê 1# x
31 ‰ 4
œ 2 Ê r ˆcos ) cos
x
È2 #
1 6
sin ) sin 16 ‰ œ 5 Ê
31 4
31 ‰ 4
sin ) sin
È3 #
œ2 Ê
È2 #
41 ‰ œ3 Ê 3 È3 # yœ3
r cos )
È2 #
r cos )
È2 #
È3 #
r ˆcos ) cos
41 3
41 ‰ 3
sin ) sin
œ 3 Ê #1 r cos ) È3 3
Ê x È 3 y œ 6 Ê y œ
È2 #
È2 #
r sin ) œ 4 Ê
x
È2 #
r sin ) œ 2
È2
È2
È3 #
r sin ) œ 3
x 2È3 1 4
sin ) sin 14 ‰ œ 4
y œ 4 Ê È2 x È2 y œ 8 Ê y œ x 4È2
5. r cos ˆ) 14 ‰ œ È2 Ê r ˆcos ) cos 14 sin ) sin 14 ‰ œ È2 Ê " r cos ) " r sin ) œ È2 Ê " x È2
" È2
y
œ È2 Ê x y œ 2 Ê y œ 2 x
6. r cos ˆ) Ê
31 ‰ 4
œ 1 Ê r ˆcos ) cos
È2 2
r cos ) Ê y œ x È 2
7. r cos ˆ)
21 ‰ 3
È2 2
È3 2
31 4
sin ) sin
31 ‰ 4
œ1
r sin ) œ 1 Ê x y œ È2
œ 3 Ê r ˆcos ) cos
Ê r cos ) 1 2
x "# y œ 5 Ê È3 x y
y œ 2 Ê È2 x È2 y œ 4 Ê y œ x 2È2
4. r cos ˆ) ˆ 14 ‰‰ œ 4 Ê r cos ˆ) 14 ‰ œ 4 Ê r ˆcos ) cos Ê
r cos ) "# r sin ) œ 5 Ê
21 3
sin ) sin " #
r sin ) œ 3 Ê x
Ê x È 3 y œ 6 Ê y œ
È3 3
È3 #
21 ‰ 3
œ3
yœ3
x 2È 3
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
668
Chapter 10 Conic Sections and Polar Coordinates
8. r cos ˆ) 13 ‰ œ 2 Ê r ˆcos ) cos Ê
1 2
r cos )
È3 2
r sin ) œ 2 Ê
Ê x È3 y œ 4 Ê y œ
È3 3
1 3
sin ) sin 13 ‰ œ 2
" #
x
x
È3 #
yœ2
4È 3 3
È 9. È2 x È2 y œ 6 Ê È2 r cos ) È2 r sin ) œ 6 Ê r Š #2 cos )
È2 #
sin )‹ œ 3 Ê r ˆcos
1 4
cos ) sin
œ 3 Ê r cos ˆ) 14 ‰ œ 3 È 10. È3 x y œ 1 Ê È3 r cos ) r sin ) œ 1 Ê r Š #3 cos )
œ
" #
Ê r cos ˆ) 16 ‰ œ
1 #
sin )‹ œ
" #
Ê r ˆcos
1 6
cos ) sin
1 6
sin )‰
" #
11. y œ 5 Ê r sin ) œ 5 Ê r sin ) œ 5 Ê r sin ()) œ 5 Ê r cos ˆ 1# ())‰ œ 5 Ê r cos ˆ) 1# ‰ œ 5 12. x œ 4 Ê r cos ) œ 4 Ê r cos ) œ 4 Ê r cos () 1) œ 4 13. r œ 2(4) cos ) œ 8 cos )
14. r œ 2(1) sin ) œ 2 sin )
15. r œ 2È2 sin )
16. r œ 2 ˆ "# ‰ cos ) œ cos )
17.
18.
19.
20.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
1 4
sin )‰
Section 10.8 Conic Sections in Polar Coordinates 21. (x 6)# y# œ 36 Ê C œ (6ß 0), a œ 6 Ê r œ 12 cos ) is the polar equation
22. (x 2)# y# œ 4 Ê C œ (2ß 0), a œ 2 Ê r œ 4 cos ) is the polar equation
23. x# (y 5)# œ 25 Ê C œ (!ß 5), a œ 5 Ê r œ 10 sin ) is the polar equation
24. x# (y 7)# œ 49 Ê C œ (!ß 7), a œ 7 Ê r œ 14 sin ) is the polar equation
25. x# 2x y# œ 0 Ê (x 1)# y# œ 1 Ê C œ (1ß 0), a œ 1 Ê r œ 2 cos ) is the polar equation
26. x# 16x y# œ 0 Ê (x 8)# y# œ 64 Ê C œ (8ß 0), a œ 8 Ê r œ 16 cos ) is the polar equation
# 27. x# y# y œ 0 Ê x# ˆy "# ‰ œ 4" Ê C œ ˆ!ß "# ‰ , a œ "# Ê r œ sin ) is the
# 28. x# y# 43 y œ 0 Ê x# ˆy 23 ‰ œ 49 Ê C œ ˆ0ß 23 ‰ , a œ 23 Ê r œ 43 sin ) is the
polar equation
polar equation
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
669
670
Chapter 10 Conic Sections and Polar Coordinates
29. e œ 1, x œ 2 Ê k œ 2 Ê r œ
2(1) 1 (1) cos )
œ
2 1cos )
30. e œ 1, y œ 2 Ê k œ 2 Ê r œ
2(1) 1 (1) sin )
œ
2 1sin )
31. e œ 5, y œ 6 Ê k œ 6 Ê r œ
6(5) 1 5 sin )
32. e œ 2, x œ 4 Ê k œ 4 Ê r œ
4(2) 1 2 cos )
33. e œ "# , x œ 1 Ê k œ 1 Ê r œ
ˆ "# ‰ (1) 1 ˆ "# ‰ cos )
35. e œ "5 , x œ 10 Ê k œ 10 Ê r œ
37. r œ
" 1 cos )
38. r œ
6 2 cos )
œ
30 15 sin )
8 12 cos )
œ
ˆ "4 ‰ (2) 1 ˆ "4 ‰ cos )
34. e œ 4" , x œ 2 Ê k œ 2 Ê r œ
36. e œ "3 , y œ 6 Ê k œ 6 Ê r œ
œ
1 2cos )
œ
ˆ "5 ‰ (10) 1 ˆ "5 ‰ sin )
ˆ "3 ‰ (6) 1 ˆ "3 ‰ sin )
œ
2 4cos )
œ
10 5sin )
6 3sin )
Ê e œ 1, k œ 1 Ê x œ 1
œ
3 1 ˆ "# ‰ cos )
Ê eœ
" #
, k œ 6 Ê x œ 6;
#
a a1 e# b œ ke Ê a ’1 ˆ "# ‰ “ œ 3 Ê
3 4
aœ3
Ê a œ 4 Ê ea œ 2
39. r œ
25 10 5 cos )
Ê eœ
" #
Ê rœ
œ
ˆ #5 ‰
1 ˆ "# ‰ cos )
, k œ 5 Ê x œ 5; a a1 e# b œ ke #
Ê a ’1 ˆ "# ‰ “ œ
40. r œ
ˆ 25 ‰ 10
5 ‰ 1 ˆ 10 cos )
4 22 cos )
Ê rœ
5 #
Ê
2 1cos )
3 4
aœ
5 #
Ê aœ
10 3
Ê ea œ
5 3
Ê e œ 1, k œ 2 Ê x œ 2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 10.8 Conic Sections in Polar Coordinates 41. r œ eœ
400 16 8 sin ) " #
Ê rœ
ˆ 400 ‰ 16
8 ‰ 1 ˆ 16 sin )
25 1 ˆ "# ‰ sin )
, k œ 50 Ê y œ 50; a a1 e# b œ ke #
Ê a ’1 ˆ "# ‰ “ œ 25 Ê Ê ea œ
42. r œ
Ê rœ
a œ 25 Ê a œ
3 4
100 3
50 3
12 3 3 sin )
Ê rœ
4 1 sin )
Ê e œ 1,
43. r œ
kœ4 Ê yœ4
44. r œ
4 2 sin )
Ê rœ
8 2 2 sin )
Ê rœ
4 1 sin )
Ê e œ 1,
k œ 4 Ê y œ 4
2 1 ˆ "# ‰ sin )
Ê eœ
" #
,kœ4 #
Ê y œ 4; a a1 e# b œ ke Ê a ’1 ˆ "# ‰ “ œ 2 Ê
3 4
aœ2 Ê aœ
8 3
Ê ea œ
4 3
45.
46.
47.
48.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
671
672
Chapter 10 Conic Sections and Polar Coordinates
49.
50.
51.
52.
53.
54.
55.
56.
57. (a) Perihelion œ a ae œ a(1 e), Aphelion œ ea a œ a(1 e) (b) Planet Perihelion Aphelion Mercury 0.3075 AU 0.4667 AU Venus 0.7184 AU 0.7282 AU Earth 0.9833 AU 1.0167 AU Mars 1.3817 AU 1.6663 AU Jupiter 4.9512 AU 5.4548 AU Saturn 9.0210 AU 10.0570 AU Uranus 18.2977 AU 20.0623 AU Neptune 29.8135 AU 30.3065 AU Pluto 29.6549 AU 49.2251 AU
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 10.8 Conic Sections in Polar Coordinates (0.3871) a1 0.2056# b 0.3707 œ 1 0.2056 1 0.2056 cos ) cos ) (0.7233) a1 0.0068# b 0.7233 Venus: r œ 1 0.0068 cos ) œ 1 0.0068 cos ) 0.0167# b 0.9997 Earth: r œ 11a10.0167 cos ) œ 1 0.0617 cos ) a1 0.0934# b 1.511 Mars: r œ (1.524) œ 1 0.0934 1 0.0934 cos ) cos ) (5.203) a1 0.0484# b 5.191 Jupiter: r œ 1 0.0484 cos ) œ 1 0.0484 cos ) a1 0.0543# b 9.511 Saturn: r œ (9.539) œ 1 0.0543 1 0.0543 cos ) cos ) (19.18) a1 0.0460# b 19.14 Uranus: r œ 1 0.0460 cos ) œ 1 0.0460 cos ) a1 0.0082# b 30.06 Neptune: r œ (30.06) œ 1 0.0082 1 0.0082 cos ) cos )
58. Mercury: r œ
59. (a) r œ 4 sin ) Ê r# œ 4r sin ) Ê x# y# œ 4y; È r œ È3 sec ) Ê r œ cos3) Ê r cos ) œ È3
(b)
#
Ê x œ È3 ; x œ È3 Ê ŠÈ3‹ y# œ 4y Ê y# 4y 3 œ 0 Ê (y 3)(y 1) œ 0 Ê y œ 3 or y œ 1. Therefore in Cartesian coordinates, the points of intersection are ŠÈ3ß 3‹ and ŠÈ3ß 1‹. In polar coordinates, 4 sin ) œ È3 sec ) Ê 4 sin ) cos ) œ È3 Ê 2 sin ) cos ) œ 21 3
Ê )œ
1 6
1 3
or
È3 #
;)œ
Ê sin 2) œ 1 6
È3 #
Ê 2) œ
Ê r œ 2, and ) œ
1 3
or
1 3
Ê r œ 2È3 Ê ˆ2ß 16 ‰ and Š2È3ß 13 ‹ are the points of intersection in polar coordinates. 60. (a) r œ 8 cos ) Ê r# œ 8r cos ) Ê x# y# œ 8x Ê x# 8x y# œ 0 Ê (x 4)# y# œ 16; r œ 2 sec ) Ê r œ cos2 ) Ê r cos ) œ 2
(b)
Ê x œ 2; x œ 2 Ê 2# 8(2) y# œ 0 Ê y# œ 12 Ê y œ „ 2È3. Therefore Š2ß „ 2È3‹
are the points of intersection in Cartesian coordinates. In polar coordinates, 8 cos ) œ 2 sec ) Ê 8 cos# ) œ 2 Ê cos# ) œ "4 Ê cos ) œ „ #" Ê ) œ 13 , 231 , 431 , or 51 3
Ê r œ 4, and ) œ 231 and 431 Ê r œ 4 Ê ˆ4ß 13 ‰ and ˆ4ß 531 ‰ are the points of intersection in polar coordinates. The points ˆ4ß 231 ‰ and ˆ4ß 431 ‰ are the same points. ;)œ
1 3
and
51 3
61. r cos ) œ 4 Ê x œ 4 Ê k œ 4: parabola Ê e œ 1 Ê r œ 62. r cos ˆ) 1# ‰ œ 2 Ê r ˆcos ) cos Ê rœ
1 #
4 1 cos )
sin ) sin 1# ‰ œ 2 Ê r sin ) œ 2 Ê y œ 2 Ê k œ 2: parabola Ê e œ 1
2 1 sin )
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
673
674
Chapter 10 Conic Sections and Polar Coordinates
63. (a) Let the ellipse be the orbit, with the Sun at one focus. rmin Then rmax œ a c and rmin œ a c Ê rrmax max rmin œ
(a c) (a c) (a c) (a c)
œ
2c 2a
œ
c a
œe
(b) Let F" , F# be the foci. Then PF" PF# œ 10 where P is any point on the ellipse. If P is a vertex, then PF" œ a c and PF# œ a c Ê (a c) (a c) œ 10 Ê 2a œ 10 Ê a œ 5. Since e œ ca we have 0.2 œ
c 5
Ê c œ 1.0 Ê the pins should be 2 inches apart. 64. e œ 0.97, Major axis œ 36.18 AU Ê a œ 18.09, Minor axis œ 9.12 AU Ê b œ 4.56 (1 AU ¸ 1.49 ‚ 10) km) (a) r œ
ke 1e cos )
œ
(b) ) œ 0 Ê r œ (c) ) œ 1 Ê r œ
(18.09) c1(0.97)# d a a1 e # b 1.07 œ 10.97 1e cos ) œ 10.97 cos ) cos ) AU 1.07 ( 10.97 ¸ 0.5431 AU ¸ 8.09 ‚ 10 km 1.07 * 10.97 ¸ 35.7 AU ¸ 5.32 ‚ 10 km
65. x# y# 2ay œ 0 Ê (r cos ))# (r sin ))# 2ar sin ) œ 0 Ê r# cos# ) r# sin# ) 2ar sin ) œ 0 Ê r# œ 2ar sin ) Ê r œ 2a sin )
66. y# œ 4ax 4a# Ê (r sin ))# œ 4ar cos ) 4a# Ê r# sin# ) œ 4ar cos ) 4a# Ê r# a1 cos# )b œ 4ar cos ) 4a# Ê r# r# cos# ) œ 4ar cos ) 4a# Ê r# œ r# cos# ) 4ar cos ) 4a# Ê r# œ (r cos ) 2a)# Ê r œ „ (r cos ) 2a) Ê r r cos ) œ 2a or 2a r r cos ) œ 2a Ê r œ 12a cos ) or r œ 1cos ) ; the equations have the same graph, which is a parabola opening to the right 67. x cos ! y sin ! œ p Ê r cos ) cos ! r sin ) sin ! œ p Ê r(cos ) cos ! sin ) sin !) œ p Ê r cos () !) œ p
#
68. ax# y# b 2ax ax# y# b a# y# œ 0 Ê Ê Ê Ê Ê Ê Ê
#
ar# b 2a(r cos )) ar# b a# (r sin ))# œ 0 r% 2ar$ cos ) a# r# sin# ) œ 0 r# cr# 2ar cos ) a# a1 cos# )bd œ 0 (assume r Á 0) r# 2ar cos ) a# a# cos# ) œ 0 ar# 2ar cos ) a# cos# )b a# œ 0 (r a cos ))# œ a# Ê r a cos ) œ „ a r œ a(1 cos )) or r œ a(1 cos ));
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 10 Practice Exercises the equations have the same graph, which is a cardioid 69 - 70. Example CAS commands: Maple: with( plots );#69 f := (r,k,e) -> k*e/(1+e*cos(theta)); elist := [3/4,1,5/4]; # (a) P1 := seq( plot( f(r,-2,e), theta=-Pi..Pi, coords=polar ), e=elist ): display( [P1], insequence=true, view=[-20..20,-20..20], title="#69(a) (Section 10.8)\nk=-2" ); P2 := seq( plot( f(r,2,e), theta=-Pi..Pi, coords=polar ), e=elist ): display( [P2], insequence=true, view=[-20..20,-20..20], title="#69(a) (Section 10.8)\nk=2" ); elist2 := [7/6,5/4,4/3,3/2,2,3,5,10,20]; # (b) P3 := seq( plot( f(r,-1,e), theta=-Pi..Pi, coords=polar ), e=elist2 ): display( [P3], insequence=true, view=[-20..20,-20..20], title="#69(b) (Section 10.8)\nk=-1, e>1" ); elist3 := [1/2,1/3,1/4,1/10,1/20]; P4 := seq( plot( f(r,-1,e), theta=-Pi..Pi, coords=polar ), e=elist3 ): display( [P4], insequence=true, title="#69(b) (Section 10.8)\nk=-1, e; t0 := sqrt(3); rr := eval( r, t=t0 ); v := map( diff, r, t ); vv := eval( v, t=t0 ); a := map( diff, v, t ); aa := eval( a, t=t0 ); s := simplify(Norm( v, 2 )) assuming t::real; ss := eval( s, t=t0 ); T := v/s; TT := vv/ss ; q1 := map( diff, simplify(T), t ): NN := simplify(eval( q1/Norm(q1,2), t=t0 )); Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
850
Chapter 13 Vector-Valued Functions and Motion in Space
BB := CrossProduct( TT, NN ); kappa := Norm(CrossProduct(vv,aa),2)/ss^3; tau := simplify( Determinant(< vv, aa, eval(map(diff,a,t),t=t0) >)/Norm(CrossProduct(vv,aa),2)^3 ); a_t := eval( diff( s, t ), t=t0 ); a_n := evalf[4]( kappa*ss^2 ); Mathematica: (assigned functions and value for t0 will vary) Clear[t, v, a, t] mag[vector_]:=Sqrt[vector.vector] Print["The position vector is ", r[t_]={t Cos[t], t Sin[t], t}] Print["The velocity vector is ", v[t_]= r'[t]] Print["The acceleration vector is ", a[t_]= v'[t]] Print["The speed is ", speed[t_]= mag[v[t]]//Simplify] Print["The unit tangent vector is ", utan[t_]= v[t]/speed[t] //Simplify] Print["The curvature is ", curv[t_]= mag[Cross[v[t],a[t]]] / speed[t]3 //Simplify] Print["The torsion is ", torsion[t_]= Det[{v[t], a[t], a'[t]}] / mag[Cross[v[t],a[t]]]2 //Simplify] Print["The unit normal vector is ", unorm[t_]= utan'[t] / mag[utan'[t]] //Simplify] Print["The unit binormal vector is ", ubinorm[t_]= Cross[utan[t],unorm[t]] //Simplify] Print["The tangential component of the acceleration is ", at[t_]=a[t].utan[t] //Simplify] Print["The normal component of the acceleration is ", an[t_]=a[t].unorm[t] //Simplify] You can evaluate any of these functions at a specified value of t. t0= Sqrt[3] {utan[t0], unorm[t0], ubinorm[t0]} N[{utan[t0], unorm[t0], ubinorm[t0]}] {curv[t0], torsion[t0]} N[{curv[t0], torsion[t0]}] {at[t0], an[t0]} N[{at[t0], an[t0]}] To verify that the tangential and normal components of the acceleration agree with the formulas in the book: at[t]== speed'[t] //Simplify an[t]==curv [t] speed[t]2 //Simplify 13.6 PLANETARY MOTION AND SATELLITES 1.
T# a$
œ
41 # GM
Ê T# œ
41 # GM
a$ Ê T# œ
41 # a6.6726‚10"" Nm# kg# b a5.975‚10#% kgb
(6,808,000 m)$
¸ 3.125 ‚ 10( sec# Ê T ¸ È3125 ‚ 10% sec# ¸ 55.90 ‚ 10# sec ¸ 93.2 min 2. e œ 0.0167 and perihelion distance œ 149,577,000 km and e œ Ê 0.0167 œ
(149,577,000,000 m)v#! a6.6726‚10"" Nm# kg# b a1.99‚10$! kgb
r! v#! GM
1
1 Ê v#! ¸ 9.03 ‚ 10) m# /sec#
Ê v! ¸ È9.03 ‚ 10) m# /sec# ¸ 3.00 ‚ 10% m/sec T# 41 # GM # $ a$ œ GM Ê a œ 41# T c"" # # #% $ a6.6726‚10 Nm kg b ˆ5.975‚10 kg‰ Ê a$ œ (5535 sec)# œ 3.094 ‚ 10#! m$ Ê a ¸ È3.094 41 # œ 6.764 ‚ 10' m ¸ 6764 km. Note that 6764 km ¸ "# a12,757 km 183 km 589 kmb.
3. 92.25 min œ 5535 sec and
‚ 10#! m$
# 4. T œ 1639 min œ 98,340 sec and mass of Mars œ 6.418 ‚ 10#$ kg Ê a$ œ GM 41# T "" # # #$ # $ ‚10 kgb (98,340 sec) œ a6.6726‚10 Nm kg b 4a6.418 ¸ 1.049 ‚ 10## m$ Ê a ¸ È1.049 ‚ 10## m$ 1#
œ 2.19 ‚ 10( m œ 21,900 km
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 13.6 Planetary Motion and Satellites 5. 2a œ diameter of Mars perigee height apogee height œ D 1499 km 35,800 km Ê 2(21,900) km œ D 37,299 km Ê D œ 6501 km 6. a œ 22,030 km œ 2.203 ‚ 10( m and T# œ Ê T# œ
#
41 a6.6720‚10"" Nm# kg# b a6.418‚10#$ kgb
41 # GM
a$
(2.203 ‚ 10( m)$ ¸ 9.856 ‚ 10* sec#
Ê T ¸ È9.856 ‚ 10) sec# ¸ 9.928 ‚ 10% sec ¸ 1655 min 7. (a) Period of the satellite œ rotational period of the Earth Ê period of the satellite œ 1436.1 min œ 86,166 sec; a$ œ
a6.6726‚10"" Nm# kg# b ˆ5.975‚10#% kg‰ (86,166 sec)#
GMT# 41 # $
Ê a$ œ 41 # $ ## È #" $ ¸ 7.4980 ‚ 10 m Ê a ¸ 74.980 ‚ 10 m ¸ 4.2168 ‚ 10( m œ 42,168 km (b) The radius of the Earth is approximately 6379 km Ê the height of the orbit is 42,168 6379 œ 35,789 km (c) Symcom 3, GOES 4, and Intelsat 5 GMT# 41 # a6.6726‚10c"" Nm# kg# b a6.418‚10#$ kgb (88,644 sec)# œ 41 # (
8. T œ 1477.4 min œ 88,644 sec Ê a$ œ œ
$ 8.524 ‚ 10#" m$ Ê a ¸ È 8.524 ‚ 10#" m$
¸ 2.043 ‚ 10 m œ 20,430 km
9. Period of the Moon œ 2.36055 ‚ 10' sec Ê a$ œ œ
a6.6726‚10c"" Nm# kg# b ˆ5.975‚10#% kg‰ (2.36055‚10' sec)# 41 # )
GMT# 41 #
$ ¸ 5.627 ‚ 10#& m$ Ê a ¸ È 5.627 ‚ 10#& m$
¸ 3.832 ‚ 10 m œ 383,200 km from the center of the Earth. 10. r œ
Ê v# œ
GM v#
11. Solar System:
T# a$
œ
É a6.6726‚10 Ê kvk œ É GM r œ
"" Nm# kg# b a5.975‚10#% kgb
41 # a6.6726‚10"" Nm# kg# b a1.99‚10$! kgb
¸ 2.97 ‚ 10"* sec# /m$ ;
#
r
¸ 1.9967 ‚ 10( r"Î# m/sec
Earth:
#
T a$
œ
41 a6.6726‚10"" Nm# kg# b a5.975‚10#% kgb
¸ 9.902 ‚ 10"% sec# /m$ ;
Moon:
T# a$
œ
41 # a6.6726‚10"" Nm# kg# b a7.354‚10## kgb
¸ 8.045 ‚ 10"# sec# /m$ ;
r! v#! GM
12. e œ
GM r
1 Ê v#! œ
GM(e 1) r!
Ê v! œ É GM(er! 1) ;
Circle: e œ 0 Ê v! œ É GM r! 2GM Ellipse: 0 e 1 Ê É GM r! v! É r!
Parabola: e œ 1 Ê v! œ É 2GM r! Hyperbola: e 1 Ê v! É 2GM r! 13. r œ
Ê v# œ
GM v#
14. ?A œ
" #
GM r
Ê v œ É GM r which is constant since G, M, and r (the radius of orbit) are constant
kr(t ?t) ‚ r(t)k Ê
œ
" #
¹ r(t ??t)t r(t) ‚ r(t)
œ
" #
¸ ddtr ‚ r(t)¸ œ
" #
?A ?t
œ
" #
?t) ‚ r(t)¹ œ ¹ r(t ?t
" #
r(t) r(t) ‚ r(t)¹ ¹ r(t ?t) ? t
r(t) ‚ r(t)¹ œ #" ¹ r(t ??t)t r(t) ‚ r(t)¹ Ê ¸r(t) ‚ ddtr ¸ œ "# kr ‚ rÞ k " ?t
dA dt
œ lim
"
?t Ä 0 #
¹ r(t ??t)t r(t) ‚ r(t)¹
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
851
852
Chapter 13 Vector-Valued Functions and Motion in Space #
# %
r v#
# %
#
! ! 15. T œ Š 2r!1va! ‹ È1 e# Ê T# œ Š 4r1# va# ‹ a1 e# b œ Š 4r1# va# ‹ ”1 Š GM 1‹ • (from Equation 32) ! ! ! !
r# v %
# %
r v#
# %
! ! œ Š 4r1# va# ‹ ’ G!# M!# 2 Š GM ‹“ œ Š 4r1# va# ‹ ’ !
!
# %
œ a41 a
!
2GM r v# 2 ‰ b Š 2r! GM! ! ‹ ˆ GM
!
2GMr! v#! r#! v%! “ G# M#
œ
ˆ41# a% ‰ a2GM r! v#! b r! G# M#
" ‰ˆ 2 ‰ # œ a41 a b ˆ 2a GM (from Equation 35) Ê T œ # %
4 1 # a$ GM
Ê
T# a$
œ
41 # GM
16. Let rAB (t) denote the vector from planet A to planet B at time t. Then rAB (t) œ rB (t) rA (t) œ [3 cos (1t) 2 cos (21t)]i [3 sin (1t) 2 sin (21t)]j œ c3 cos (1t) 2 acos# (1t) sin# (1t)bd i [3 sin (1t) 4 sin (1t) cos (1t)]j œ c3 cos (1t) 4 cos# (1t) 2d i [(3 4 cos (1t)) sin (1t)]j Ê parametric equations for the path are x(t) œ 2 [3 4 cos (1t)] cos (1t) and y(t) œ [3 4 cos (1t)] sin (1t) 17. The graph of the path of planet B is the limacon ¸ at the right.
18. (i) (ii) (iii) (iv) (v)
Perihelion is the time t such that kr(t)k is a minimum. Aphelion is the time t such that kr(t)k is a maximum. Equinox is the time t such that r(t) † w œ 0 . Summer solstice is the time t such that the angle between r(t) and w is a maximum. Winter solstice is the time t such that the angle between r(t) and w is a minimum.
CHAPTER 13 PRACTICE EXERCISES 1. r(t) œ (4 cos t)i ŠÈ2 sin t‹ j Ê x œ 4 cos t and y œ È2 sin t Ê
x# 16
y# #
œ 1;
v œ (4 sin t)i ŠÈ2 cos t‹ j and a œ (4 cos t)i ŠÈ2 sin t‹ j ; r(0) œ 4 i , v(0) œ È2j , a(0) œ 4i ; r ˆ 14 ‰ œ 2È2i j , v ˆ 14 ‰ œ 2È2i j , a ˆ 1 ‰ œ 2È2i j ; kvk œ È16 sin# t 2 cos# t 4
Ê aT œ
d dt
kvk œ
at t œ 14 : aT œ
14 sin t cos t È16 sin# t2 cos# t
7 È 8 1
œ
7 3
; at t œ 0: aT œ 0, aN œ Ékak# 0 œ 4, , œ
, aN œ É9
49 9
œ
4È 2 3
,,œ
aN kv k #
œ
aN kv k #
œ
4 2
œ 2;
4È 2 27
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 13 Practice Exercises 2. r(t) œ ŠÈ3 sec t‹ i ŠÈ3 tan t‹ j Ê x œ È3 sec t and y œ È3 tan t Ê
x# 3
y# 3
853
œ sec# t tan# t œ 1;
Ê x# y# œ 3; v œ ŠÈ3 sec t tan t‹ i ŠÈ3 sec# t‹ j and a œ ŠÈ3 sec t tan# t È3 sec$ t‹ i Š2È3 sec# t tan t‹ j ; r(0) œ È3i , v(0) œ È3j , a(0) œ È3i ; kvk œ È3 sec# t tan# t 3 sec% t Ê aT œ
d dt
kvk œ
6 sec# t tan$ t 18 sec% t tan t 2È3 sec# t tan# t 3 sec% t
;
at t œ 0: aT œ 0, aN œ Ékak# 0 œ È3, ,œ 3. r œ
œ
È3 3
œ
" È1 t#
i
t È1 t#
aN kv k #
" È3
j Ê v œ t a1 t# b
$Î#
i a 1 t# b
#
#
Ê kvk œ Ê’t a1 t# b$Î# “ ’a1 t# b$Î# “ œ d kv k dt
œ0 Ê
2t a1 t# b#
œ 0 Ê t œ 0. For t 0,
" 1 t#
2t a1 t# b#
$Î#
j
. We want to maximize kvk :
0; for t 0,
2t a1 t# b#
d kv k dt
œ
2t a1 t# b#
and
0 Ê kvk max occurs when
t œ 0 Ê kvk max œ 1 4. r œ aet cos tb i aet sin tb j Ê v œ aet cos t et sin tb i aet sin t et cos tb j Ê a œ aet cos t et sin t et sin t et cos tb i aet sin t et cos t et cos t et sin tb j œ a2et sin tb i a2et cos tb j . Let ) be the angle between r and a . Then ) œ cos" Š krrk†kaak ‹ œ cos"
2e2t sin t cos t2e2t sin t cos t Éaet cos tb# aet sin tb# Éa2et sin tb# a2et cos tb#
â âi â 5. v œ 3i 4j and a œ 5i 15j Ê v ‚ a œ â 3 â â5 Ê ,œ 6. , œ
kv ‚ a k kv k $
kyww k $Î# 1 ayw b# ‘
œ ex a1 e2x b d, dx
œ
25 5$
œ
$Î#
&Î#
œ 0 Ê a1 2e2x b œ 0 Ê e2x œ
maximum at the point Š ln È2ß
x# y# œ 1, 2x
dx dt
dx dt
i
2y
dy dt
dy dt
d, dx
Ê
3e3x a1 e2x b
7. r œ xi yj Ê v œ
for all t
â kâ â 0 â œ 25k Ê kv ‚ ak œ 25; kvk œ È3# 4# œ 5 â 0â
j 4 "5
" 5
œ ex a1 e2x b $Î#
1 #
œ cos" Š 2e02t ‹ œ cos" 0 œ
" #
œ ex a1 e2x b
œ ex a1 e2x b
$Î#
ex ’ #3 a1 e2x b
&Î#
&Î#
a2e2x b“ &Î#
ca1 e2x b 3e2x d œ ex a1 e2x b a1 2e2x b ; Ê 2x œ ln 2 Ê x œ "# ln 2 œ ln È2 Ê y œ È"2 ; therefore , is at a
" È2 ‹
j and v † i œ y Ê
œ0 Ê
dy dt
œ
x dx y dt
dx dt
œ y. Since the particle moves around the unit circle
Ê
dy dt
œ yx (y) œ x. Since
dx dt
œ y and
dy dt
œ x, we have
v œ yi xj Ê at (1ß 0), v œ j and the motion is clockwise. dy dy " # dx # dx dt œ 3x dt Ê dt œ 3 x dt . If r œ xi yj , where x and y are differentiable functions of dy dy dx " # dx " # then v œ dx dt i dt j. Hence v † i œ 4 Ê dt œ 4 and v † j œ dt œ 3 x dt œ 3 (3) (4) œ 12 at (3ß 3). Also, # # # # # ‰# ˆ 3" x# ‰ ddt#x . Hence a † i œ 2 Ê ddt#x œ 2 and a œ ddt#x i ddt#y j and ddt#y œ ˆ 23 x‰ ˆ dx dt # a † j œ ddt#y œ 23 (3)(4)# "3 (3)# (2) œ 26 at the point (xß y) œ (3ß 3).
8. 9y œ x$ Ê 9
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
t,
854 9.
Chapter 13 Vector-Valued Functions and Motion in Space dr dt
orthogonal to r Ê 0 œ
†rœ
dr dt
" dr # dt
† r "# r †
œ
dr dt
#
" d # dt #
(r † r) Ê r † r œ K, a constant. If r œ xi yj , where
x and y are differentiable functions of t, then r † r œ x y Ê x# y# œ K, which is the equation of a circle centered at the origin. 10. (b) v œ (1 1 cos 1t)i (1 sin 1t)j Ê a œ a1# sin 1tb i a1# cos 1tb j ; v(0) œ 0 and a(0) œ 1# j ; v(1) œ 21i and a(1) œ 1# j ; v(2) œ 0 and a(2) œ 1# j ; v(3) œ 21i and a(3) œ 1# j
(c) Forward speed at the topmost point is kv(1)k œ kv(3)k œ 21 ft/sec; since the circle makes
" #
revolution per
second, the center moves 1 ft parallel to the x-axis each second Ê the forward speed of C is 1 ft/sec. 11. y œ y! (v! sin !)t "# gt# Ê y œ 6.5 (44 ft/sec)(sin 45°)(3 sec) "# a32 ft/sec# b (3 sec)# œ 6.5 66È2 144 ¸ 44.16 ft Ê the shot put is on the ground. Now, y œ 0 Ê 6.5 22È2t 16t# œ 0 Ê t ¸ 2.13 sec (the positive root) Ê x ¸ (44 ft/sec)(cos 45°)(2.13 sec) ¸ 66.27 ft or about 66 ft, 3 in. from the stopboard 12. ymax œ y!
(v! sin !)# #g
œ 7 ft
[(80 ft/sec)(sin 45°)]# (2) a32 ft/sec# b
¸ 57 ft (v! sin !)t "# gt# (v sin !) " gt y œ ! v! cos ! # x œ (v! cos !)t 2v! sin ! 2v! cos ! tan 9 , which is the time when g
13. x œ (v! cos !)t and y œ (v! sin !)t "# gt# Ê tan 9 œ Ê v! cos ! tan 9 œ v! sin ! "# gt Ê t œ
the golf ball
hits the upward slope. At this time x œ (v! cos !) Š 2v! sin ! 2vg ! cos ! tan 9 ‹ œ Š 2g ‹ av#! sin ! cos ! v#! cos# ! tan 9b . Now OR œ
x cos 9
Ê OR œ Š g2 ‹ Š
v#! sin ! cos ! v#! cos# ! tan 9 ‹ cos 9
œŠ
2v#! cos ! sin ! ‹ Š cos g 9
œŠ
2v#! cos ! 9 cos ! sin 9 ‹ Š sin ! cos cos ‹ #9 g
œŠ
2v#! cos ! g cos# 9 ‹ [sin (!
cos ! tan 9 cos 9 ‹
9)]. The distance OR is maximized
when x is maximized:
dx d!
œŠ
2v#! g ‹(cos
Ê cot 2! œ tan (9) Ê 2! œ 14. R œ
v#! g
1 #
2! sin 2! tan 9) œ 0 Ê (cos 2! sin 2! tan 9) œ 0 Ê cot 2! tan 9 œ 0
9 Ê !œ
9 #
1 4 #
ft) a32 ft/sec b sin 2! Ê v! œ É sinRg2! ; for 4325 yards: 4325 yards œ 12,975 ft Ê v! œ É (12,975(sin 90°) #
ft) a32 ft/sec b ¸ 644 ft/sec; for 4752 yards: 4752 yards œ 14,256 ft Ê v! œ É (14,256(sin ¸ 675 ft/sec 90°)
15. (a) R œ
v#! g
v#
# # # ! sin 2! Ê 109.5 ft œ Š 32 ft/sec Ê v! œ È3504 ft# /sec# # ‹ (sin 90°) Ê v! œ 3504 ft /sec
¸ 59.19 ft/sec (b) x œ (v! cos !)t and y œ 4 (v! sin !)t "# gt# ; when the cork hits the ground, x œ 177.75 ft and y œ 0 Ê 177.75 œ Šv! Ê v! œ
(177.75)È2 t
" È2 ‹ t
œ
and 0 œ 4 Šv!
4(177.75)È2 È181.75
" È2 ‹ t
16t# Ê 16t# œ 4 177.75 Ê t œ
È181.75 4
¸ 74.58 ft/sec
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 13 Practice Exercises
855
5 16. (a) x œ v! (cos 40°)t and y œ 6.5 v! (sin 40°)t "# gt# œ 6.5 v! (sin 40°)t 16t# ; x œ 262 12 ft and y œ 0 ft 5 Ê 262 12 œ v! (cos 40°)t or v! œ
262.4167 # # and 0 œ 6.5 ’ (cos 40°)t “ (sin 40°)t 16t Ê t œ 14.1684
262.4167 (cos 40°)t
Ê t ¸ 3.764 sec. Therefore, 262.4167 ¸ v! (cos 40°)(3.764 sec) Ê v! ¸ #
(v! sin !) 2g
(b) ymax œ y!
a(91)(sin 40°)b (2)(32)
¸ 6.5
2
262.4167 (cos 40°)(3.764 sec)
Ê v! ¸ 91 ft/sec
¸ 60 ft
#
#
17. x# œ av!# cos# !b t# and ˆy "# gt# ‰ œ av!# sin# !b t# Ê x# ˆy "# gt# ‰ œ v!# t# Þ ÞÞ Þ ÞÞ ÞÞ# ÞÞ# ÞÞ# ÞÞ# ÞÞ# axÞ ÞÞx yÞ ÞÞyb# x x y y Ê x y s œ x y xÞ # yÞ # Þ Þ # # Èx y ÞÞ# ÞÞ# Þ # Þ # Þ # ÞÞ# Þ ÞÞ Þ ÞÞ Þ # ÞÞ# Þ ÞÞ Þ ÞÞ Þ ÞÞ Þ ÞÞ Þ ÞÞ Þ ÞÞ ax y b ax y b ax x 2x x y y y y b ax y y x b# x# y# y# x# 2x x y y œ œ œ Þ# Þ# Þ# Þ# Þ Þ x y x y x# y# Þ ÞÞ Þ ÞÞ Þ # Þ # $Î# Þ Þ # # kx y y x k ax y b ÞÞ ÞÞ ÞÞ x y Ê È x# y# s # œ È Þ # Þ # Ê ÈÞÞ# ÞÞ# ÞÞ# œ kxÞ ÞÞy yÞ ÞÞxk œ ," œ 3 x y x y s
ÞÞ 18. s œ
d dt
ÈxÞ # yÞ # œ
19. r(t) œ ’'0 cos ˆ "# 1)# ‰ d)“ i ’'0 sin ˆ "# 1)# ‰ d)“ j Ê v(t) œ cos Š 1#t ‹ i sin Š 1#t â i j â â # 1t# # # â sin Š 1#t ‹ a(t) œ 1t sin Š 1#t ‹ i 1t cos Š 1#t ‹ j Ê v ‚ a œ â cos Š # ‹ â â 1t sin Š 1t# ‹ 1t cos Š 1t# ‹ â # # t
t
œ 1 tk Ê , œ 20. s œ a) Ê ) œ
kv ‚ a k kv k $ s a
#
œ 1t; kv(t)k œ
Ê 9œ
s a
1 #
ds dt
Ê
#
‹ j Ê kvk œ 1; k ââ â 0 ââ â 0 ââ
œ 1 Ê s œ t C; r(0) œ 0 Ê s(0) œ 0 Ê C œ 0 Ê , œ 1s d9 ds
œ
" a
Ê , œ ¸ "a ¸ œ
" a
since a 0
21. r œ (2 cos t)i (2 sin t)j t# k Ê v œ (2 sin t)i (2 cos t)j 2tk Ê kvk œ È(2 sin t)# (2 cos t)# (2t)# œ 2È1 t# Ê Length œ '0 2È1 t# dt œ ’tÈ1 t# ln ¹t È1 t# ¹“ 1Î4
1Î% !
œ
1 4
É1
1# 16
ln Š 14 É1
22. r œ (3 cos t)i (3 sin t)j 2t$Î# k Ê v œ (3 sin t)i (3 cos t)j 3t"Î# k
1# 16 ‹
$ # Ê kvk œ É(3 sin t)# (3 cos t)# a3t"Î# b œ È9 9t œ 3È1 t Ê Length œ '0 3È1 t dt œ 2(1 t)$Î# ‘ ! 3
œ 14 23. r œ
4 9
(1 t)$Î# i 49 (1 t)$Î# j 3" tk Ê v œ #
2 3
(1 t)"Î# i 32 (1 t)"Î# j 3" k
#
#
Ê kvk œ É 23 (1 t)"Î# ‘ 23 (1 t)"Î# ‘ ˆ 3" ‰ œ 1 Ê T œ i 23 j 3" k ;
" 3
(1 t)"Î# i â â â Ê N(0) œ È" i È" j ; B(0) œ T(0) ‚ N(0) œ ââ 2 2 â â Ê T(0) œ
2 3
dT dt
œ
2 3
(1 t)"Î# i 23 (1 t)"Î# j 3" k
" 3
(1 t)"Î# j Ê ddtT (0) œ 3" i 3" j Ê ¸ ddtT (0)¸ œ â i j kâ 2 " â " " 4 â 23 3 3 â œ È i È j È k; 3 2 3 2 3 2 " " 0 ââ È2 È2
a œ "3 (1 t)"Î# i 3" (1 t)"Î# j Ê a(0) œ 3" i 3" j and v(0) œ 32 i 32 j 3" k Ê v(0) ‚ a(0) â i j k ââ È â Š 32 ‹ È2 È2 â 2 2 "â kv ‚a k " " 4 i j k k v a k œâ 3 œ Ê ‚ œ Ê , (0) œ œ â 3 3 9 9 9 3 1$ œ 3 ; kv k $ â " â " â 3 0â 3 Þ Þ a œ "6 (1 t)$Î# i "6 (1 t)$Î# j Ê a(0) œ 6" i 6" j Ê 7 (0) œ
â 2 â 3 â â " â 3 â " â 6
23 " 3 " 6
kv ‚a k #
â â â 0 ââ 0 ââ " 3
œ
2 ‰ ˆ "3 ‰ ˆ 18
Š
È2 ‹# œ 3
t œ 0 Ê ˆ 49 ß 49 ß 0‰ is the point on the curve
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
" 6
;
È2 3
856
Chapter 13 Vector-Valued Functions and Motion in Space
24. r œ aet sin 2tb i aet cos 2tb j 2et k Ê v œ aet sin 2t 2et cos 2tb i aet cos 2t 2et sin 2tb j 2et k Ê kvk œ Éaet sin 2t 2et cos 2tb# aet cos 2t 2et sin 2tb# a2et b# œ 3et Ê T œ œ ˆ "3 sin 2t dT dt
2 3
cos 2t‰ i ˆ 3" cos 2t
œ ˆ 23 cos 2t
Ê N(0) œ
2 3
sin 2t‰ j 23 k Ê T(0) œ
sin 2t‰ i ˆ 23 sin 2t
4 3
4 3
cos 2t‰ j Ê
dT dt
È
Š2 3 5‹
i 3" j 32 k ;
i 43 j Ê ¸ ddtT (0)¸ œ 23 È5 â j kâ 1 2 â 4 2 5 â 3 3 â œ È i È j È k; 3 5 3 5 3 5 2 â È5 0 â
(0) œ
â â i â 2 2 " œ È5 i È5 j ; B(0) œ T(0) ‚ N(0) œ ââ 3 â " â È5
ˆ 23 i 43 j‰
2 3 2 3
v kv k
a œ a4et cos 2t 3et sin 2tb i a3et cos 2t 4et sin 2tb j 2et k Ê a(0) œ 4i 3j 2k and v(0) œ 2i j 2k â â j kâ âi â â Ê v(0) ‚ a(0) œ â 2 " 2 â œ 8i 4j 10k Ê kv ‚ ak œ È64 16 100 œ 6È5 and kv(0)k œ 3 â â â 4 3 2 â Ê ,(0) œ
6È 5 3$
œ
2È 5 9 t
â â 2 â â 4 â â 2
1 3 11
; Þ t a œ a4e cos 2t 8e sin 2t 3et sin 2t 6et cos 2tb i a3et cos 2t 6et sin 2t 4et sin 2t 8et cos 2tb j 2et k Þ œ a2et cos 2t 11et sin 2tb i a11 et cos 2t 2et sin 2tb j 2et k Ê a(0) œ 2i 11j 2k Ê 7 (0) œ
kv ‚ a k #
â 2â â 2â â 2â
œ
80 180
œ 94 ; t œ 0 Ê (!ß "ß 2) is on the curve
25. r œ ti "# e2t j Ê v œ i e2t j Ê kvk œ È1 e4t Ê T œ dT dt
œ
2 e ˆ1 e4t ‰$Î# 4t
i
2t
2e ˆ1 e4t ‰$Î#
j Ê
dT dt
(ln 2) œ
32 17È17
i
8 17È17
" È1 e4t
i
e2t È1 e4t
j Ê T (ln 2) œ
j Ê N (ln 2) œ È417 i
" È17
" È17
i
4 È17
j;
j;
â i j k ââ â â " 4 0 ââ œ k ; a œ 2e2t j Ê a(ln 2) œ 8j and v(ln 2) œ i 4j È17 B (ln 2) œ T(ln 2) ‚ N(ln 2) œ ââ È17 â " â 4 â â È17 È17 0 â â â â i j kâ â â Þ 8 Ê v(ln 2) ‚ a(ln 2) œ â " 4 0 â œ 8k Ê kv ‚ ak œ 8 and kv(ln 2)k œ È17 Ê ,(ln 2) œ 17È ; a œ 4e2t j 17 â â â0 8 0â Þ Ê a(ln 2) œ 16j Ê 7 (ln 2) œ
â â1 â â0 â â0
4 8 16 kv ‚a k #
â 0â â 0â â 0â
œ 0; t œ ln 2 Ê (ln 2ß 2ß 0) is on the curve
26. r œ (3 cosh 2t)i (3 sinh 2t)j 6tk Ê v œ (6 sinh 2t)i (6 cosh 2t)j 6k Ê kvk œ È36 sinh# 2t 36 cosh# 2t 36 œ 6È2 cosh 2t Ê T œ kvvk œ Š È" tanh 2t‹ i 2
Ê T(ln 2) œ #
15 17È2
i
" È2
j
8 17È2
k;
8 ‰ 8 ‰ ˆ 15 ‰ œ Š È22 ‹ ˆ 17 i Š È22 ‹ˆ 17 17 k œ
dT dt
œ
Š È22
sech 2t‹ i
Š È22
sech 2t tanh 2t‹ k Ê
j Š È" sech 2t‹ k
dT dt
(ln 2)
2
#
#
È
8 2 128 240 k Ê ¸ ddtT (ln 2)¸ œ ÊŠ 289 È2 ‹ Š 289È2 ‹ œ 17 â â j k â â i â â " 15 8 " 8 15 8 Ê N(ln 2) œ 17 i 17 k ; B(ln 2) œ T(ln 2) ‚ N(ln 2) œ ââ 17È2 È2 17È2 ââ œ 1715 È2 i È2 j 17È2 k ; â 8 â 0 15 â 17 17 â 17 ‰ 15 ‰ 51 ˆ ˆ a œ (12 cosh 2t)i (12 sinh 2t)j Ê a(ln 2) œ 12 8 i 12 8 j œ # i 45 # j and â i j k ââ â â 45 51 45 51 ‰ ˆ 17 ‰ 6 ââ v(ln 2) œ 6 ˆ 15 4 8 i 6 8 j 6k œ 4 i 4 j 6k Ê v(ln 2) ‚ a(ln 2) œ ââ 4 â 45 â 51 0â 2 # 128 289È2
i
#
" È2
240 289È2
œ 135i 153j 72k Ê kv ‚ ak œ 153È2 and kv(ln 2)k œ
51 4
È2 Ê ,(ln 2) œ
153È2 $ Š 51 È2‹
œ
4
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
32 867
;
Chapter 13 Practice Exercises
Þ Þ a œ (24 sinh 2t)i (24 cosh 2t)j Ê a(ln 2) œ 45i 51j Ê 7 (ln 2) œ
â 45 â 4 â 5" â 2 â â 45
51 4 45 2
51 kv ‚a k #
â 6â â 0â â 0â
œ
32 867
857
;
45 ‰ t œ ln 2 Ê ˆ 51 8 ß 8 ß 6 ln 2 is on the curve
27. r œ a2 3t 3t# b i a4t 4t# b j (6 cos t)k Ê v œ (3 6t)i (4 8t)j (6 sin t)k Ê kvk œ È(3 6t)# (4 8t)# (6 sin t)# œ È25 100t 100t# 36 sin# t "Î#
" #
a25 100t 100t# 36 sin# tb (100 200t 72 sin t cos t) Ê aT (0) œ ddtkvk (0) œ 10; a œ 6i 8j (6 cos t)k Ê kak œ È6# 8# (6 cos t)# œ È100 36 cos# t Ê ka(0)k œ È136 Ê
d kv k dt
œ
Ê aN œ Ékak# a#T œ È136 10# œ È36 œ 6 Ê a(0) œ 10T 6N 28. r œ (2 t)i at 2t# b j a1 t# b k Ê v œ i (1 4t)j 2tk Ê kvk œ È1# (1 4t)# (2t)# "Î# œ È2 8t 20t# Ê ddtkvk œ "# a2 8t 20t# b (8 40t) Ê aT œ ddtkvk (0) œ 2È2; a œ 4j 2k #
Ê kak œ È4# 2# œ È20 Ê aN œ Ékak# aT# œ Ê20 Š2È2‹ œ È12 œ 2È3 Ê a(0) œ 2È2T 2È3N 29. r œ (sin t)i ŠÈ2 cos t‹ j (sin t)k Ê v œ (cos t)i ŠÈ2 sin t‹ j (cos t)k #
Ê kvk œ Ê(cos t)# ŠÈ2 sin t‹ (cos t)# œ È2 Ê T œ
v kv k
œ Š È"2 cos t‹ i (sin t)j Š È"2 cos t‹ k ; #
#
œ Š È"2 sin t‹ i (cos t)j Š È"2 sin t‹ k Ê ¸ ddtT ¸ œ ÊŠ È"2 sin t‹ ( cos t)# Š È"2 sin t‹ œ 1 â â i j k â â â " cos t sin t â " ˆ ddtT ‰ cos t " " â â È2 Ê N œ ¸ dT ¸ œ Š È2 sin t‹ i (cos t)j Š È2 sin t‹ k ; B œ T ‚ N œ â È2 â dt â " sin t cos t " sin t â â È2 â È2 â i â j k â â â â œ È"2 i È"2 k ; a œ ( sin t)i ŠÈ2 cos t‹ j (sin t)k Ê v ‚ a œ â cos t È2 sin t cos t â â â â sin t È2 cos t sin t â Þ œ È2 i È2 k Ê kv ‚ ak œ È4 œ 2 Ê , œ kv‚$ak œ 2 $ œ " ; a œ ( cos t)i ŠÈ2 sin t‹ j (cos t)k
dT dt
kv k
Ê 7œ
â â cos t â â sin t â â â cos t
â cos t ââ È2 sin t È2 cos t sin t ââ È2 sin t cos t ââ kv ‚a k #
œ
ŠÈ2‹
È2
(cos t) ŠÈ2‹ ŠÈ2 sin t‹ (0) (cos t) ŠÈ2‹ 4
œ0
30. r œ i (5 cos t)j (3 sin t)k Ê v œ (5 sin t)j (3 cos t)k Ê a œ (5 cos t)j (3 sin t)k Ê v † a œ 25 sin t cos t 9 sin t cos t œ 16 sin t cos t; v † a œ 0 Ê 16 sin t cos t œ 0 Ê sin t œ 0 or cos t œ 0 Ê t œ 0, 1# or 1 31. r œ 2i ˆ4 sin #t ‰ j ˆ3 1t ‰ k Ê 0 œ r † (i j) œ 2(1) ˆ4 sin #t ‰ (1) Ê 0 œ 2 4 sin Ê tœ
1 3
t #
Ê sin
t #
œ
" #
Ê
t #
œ
(for the first time)
32. r(t) œ ti t# j t$ k Ê v œ i 2tj 3t# k Ê kvk œ È1 4t# 9t% Ê kv(1)k œ È14 Ê T(1) œ È"14 i È214 j È314 k , which is normal to the normal plane Ê
" È14
(x 1)
2 È14
(y 1)
3 È14
(z 1) œ 0 or x 2y 3z œ 6 is an equation of the normal plane. Next we
calculate N(1) which is normal to the rectifying plane. Now, a œ 2j 6tk Ê a(1) œ 2j 6k Ê v(1) ‚ a(1)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
1 6
858
Chapter 13 Vector-Valued Functions and Motion in Space â âi â œ â" â â0 œ
" #
œ
22 È14
â j kâ â 2 3 â œ 6i 6j 2k Ê kv(1) ‚ a(1)k œ È76 Ê ,(1) œ â 2 6â
a1 4t# 9t% b
"Î#
2j3k Š iÈ ‹ 14
a8t 36t$ b¹
È19 7È14
#
œ
tœ1
22 È14
, so a œ
ŠÈ14‹ N Ê N œ
È14 2È19
d# s dt#
È76 $ È Š 14‹
œ
È19 7È14
;
ds dt
œ kv(t)k Ê
d# s dt# ¹ tœ1
#
‰ N Ê 2j 6k T , ˆ ds dt
8 9 ‰ 11 8 9 ˆ 11 7 i 7 j 7 k Ê 7 (x 1) 7 (y 1) 7 (z 1)
œ 0 or 11x 8y 9z œ 10 is an equation of the rectifying plane. Finally, B(1) œ T(1) ‚ N(1) â â j kâ â i È14 â â 2 3 â œ È" (3i 3j k) Ê 3(x 1) 3(y 1) (z 1) œ 0 or 3x 3y z œ Š 2È19 ‹ Š È" ‹ ˆ 7" ‰ â " 19 14 â â â 11 8 9 â œ 1 is an equation of the osculating plane. " ‰ 33. r œ et i (sin t)j ln (1 t)k Ê v œ et i (cos t)j ˆ 1 t k Ê v(0) œ i j k ; r(0) œ i Ê (1ß 0ß 0) is on the line
Ê x œ 1 t, y œ t, and z œ t are parametric equations of the line
34. r œ ŠÈ2 cos t‹ i ŠÈ2 sin t‹ j tk Ê v œ ŠÈ2 sin t‹ i ŠÈ2 cos t‹ j k Ê v ˆ 14 ‰ œ ŠÈ2 sin 14 ‹ i ŠÈ2 cos 14 ‹ j k œ i j k is a vector tangent to the helix when t œ is parallel to v ˆ 14 ‰ ; also r ˆ 14 ‰ œ ŠÈ2 cos 14 ‹ i ŠÈ2 sin 14 ‹ j Ê x œ 1 t, y œ 1 t, and z œ 35. (a) ?SOT ¸ ?TOD Ê Ê y! œ
6380# 6817
DO OT
œ
1 4
1 4
1 4
Ê the tangent line
k Ê the point ˆ1ß 1ß 14 ‰ is on the line
t are parametric equations of the line
OT SO
Ê
y! 6380
œ
6380 6380437
Ê y! ¸ 5971 km;
(b) VA œ '5971 21x Ê1 Š dx dy ‹ dy #
6380
6380 œ 21'5971 È6380# y# Š È6380 # y# ‹ dy 6817
œ 21 '5971 6380 dy œ 21 c6380yd ')"( &*(" 6817
œ 16,395,469 km# ¸ 1.639 ‚ 10( km# ; (c) percentage visible ¸
16,395,469 km# 41(6380 km)#
¸ 3.21%
CHAPTER 13 ADDITIONAL AND ADVANCED EXERCISES 1. (a) The velocity of the boat at (xß y) relative to land is the sum of the velocity due to the rower and the " velocity of the river, or v œ 250 (y 50)# 10‘ i 20j . Now, dy dt œ 20 Ê y œ 20t c; y(0) œ 100 " Ê c œ 100 Ê y œ 20t 100 Ê v œ 250 (20t 50)# 10‘ i 20j œ ˆ 85 t# 8t‰ i 20j 8 $ Ê r(t) œ ˆ 15 t 4t# ‰ i 20tj C" ; r(0) œ 0i 100j Ê 100j œ C" Ê r(t) 8 $ œ ˆ 15 t 4t# ‰ i (100 20t) j
(b) The boat reaches the shore when y œ 0 Ê 0 œ 20t 100 from part (a) Ê t œ 5 8 100 ‰ Ê r(5) œ ˆ 15 † 125 4 † 25‰ i (100 20 † 5)j œ ˆ 200 3 100 i œ 3 i ; the distance downstream is therefore
100 3
m
2. (a) Let ai bj be the velocity of the boat. The velocity of the boat relative to an observer on the bank of the river is v œ ai ’b
3x(20 x) “j. 100
x œ at Ê v œ ai ’b
The distance x of the boat as it crosses the river is related to time by
3at(20 at) “j 100
œ ai Šb
3a# t# 60at ‹j 100
Ê r(t) œ ati Šbt
a# t$ 100
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
30at# 100 ‹ j
C;
Chapter 13 Additional and Advanced Exercises a# t$ 30at# ‹j. 100
r(0) œ 0i 0j Ê C œ 0 Ê r(t) œ ati Šbt Ê 20 œ at Ê t œ œ
2000b 8000 12,000 100a
#
ˆ 20 ‰$ a
The boat reaches the shore when x œ 20
‰ 30a ˆ 20 a 100
#
$
#
(20) 30(20) œ 20b a 100a # # È È Ê b œ 2; the speed of the boat is 20 œ kvk œ a b œ Èa# 4 Ê a# œ 16 20 a
‰ and y œ 0 Ê 0 œ b ˆ 20 a
a
Ê a œ 4; thus, v œ 4i 2j is the velocity of the boat a# t$ 30at# ‹j 100
(b) r(t) œ ati Šbt
$
(c) x œ 4t and y œ 2t œ œ
16t$ 100
œ 4ti Š2t
120t# 100 ‹ j
by part (a), where 0 Ÿ t Ÿ 5
#
16t 120t 100 100 4 $ 6 # 2 # 25 t 5 t 2t œ 25 t a2t 15t 25b 2 25 t(2t 5)(t 5), which is the graph of
the cubic displayed here
3. (a) r()) œ (a cos ))i (a sin ))j b)k Ê œ Èa# b# (b)
d) dt
d) dt
Ê
) œ É a#2gb b# Ê
d) dt
d) È)
dr dt
œ [(a sin ))i (a cos ))j bk]
) É a#2gb œ É a#2gz b# œ b# Ê
gbt# 2 aa # b # b
; z œ b) Ê z œ
œ [(a sin ))i (a cos ))j bk]
d) dt
; kvk œ È2gz œ ¸ ddtr ¸
œ É a#41gbb# œ 2É a#1gbb#
gb# t# 2 aa # b # b
œ [(a sin ))i (a cos ))j bk] Š a# gbt b# ‹ , from part (b)
i (a cos ))j bk Ê v(t) œ ’ (a sin ))È “ Š Èagbt ‹œ # b# a# b# d# r dt#
d) ¸ dt )œ#1
d) dt
"Î# œ É a#2gb œ É a#2gb b# dt Ê 2) b# t C; t œ 0 Ê ) œ 0 Ê C œ 0
Ê 2)"Î# œ É a#2gb b# t Ê ) œ (c) v(t) œ
dr dt
gbt È a# b #
T;
#
œ [(a cos ))i (a sin ))j] ˆ ddt) ‰ [(a sin ))i (a cos ))j bk] #
d# ) dt#
gb œ Š a# gbt b# ‹ [(a cos ))i (a sin ))j] [(a sin ))i (a cos ))j bk] Š a# b# ‹ #
i (a cos ))j bk œ ’ (a sin ))È “ Š Èa#gb b# ‹ a Š a# gbt b# ‹ [( cos ))i (sin ))j] a# b#
œ
gb È a# b#
#
T a Š a# gbt b# ‹ N (there is no component in the direction of B).
4. (a) r()) œ (a) cos ))i (a) sin ))j b)k Ê
dr dt
# "Î#
kvk œ È2gz œ ¸ ddtr ¸ œ aa# a# )# b b (b) s œ '0 kvk dt œ '0 aa# a# )# b# b t
t
œ '0 aÉ a )
#
b# a#
(1 e)r! 1 e cos )
Ê
dr d)
ˆ ddt) ‰ Ê
d) dt
œ
œ
c# #
)
ln ¹u Èc# u# ¹“ œ
(1 e)r! (e sin )) (1 e cos ))#
!
;
dr d)
œ0 Ê
Ê sin ) œ 0 Ê ) œ 0 or 1. Note that
dr d)
a #
;
È a# a# ) # b#
dt œ '0 aa# a# )# b# b
u# du œ a '0 Èc# u# du, where c œ
d) dt
È2gb)
t
)
Ê s œ a ’ u# Èc# u# 5. r œ
"Î# d) dt
œ [(a cos ) a) sin ))i (a sin ) a) cos ))j bk]
"Î#
d) œ '0 aa# a# u# b# b )
"Î#
du
È a# b# ka k
Š)Èc# )# c# ln ¹) Èc# )# ¹ c# ln c‹
(1 e)r! (e sin )) (1 e cos ))#
œ 0 Ê (1 e)r! (e sin )) œ 0
0 when sin ) 0 and
dr d)
0 when sin ) 0. Since sin ) 0 on
1 ) 0 and sin ) 0 on 0 ) 1, r is a minimum when ) œ 0 and r(0) œ
(1 e)r! 1 e cos 0
œ r!
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
859
860
Chapter 13 Vector-Valued Functions and Motion in Space
6. (a) f(x) œ x 1
" #
sin x œ 0 Ê f(0) œ 1 and f(2) œ 2 1
" #
sin 2
" #
since ksin 2k Ÿ 1; since f is continuous
on [0ß 2], the Intermediate Value Theorem implies there is a root between 0 and 2 (b) Root ¸ 1.4987011335179 7. (a) v œ
dx dt
i
v†iœ Ê
j and v œ
dr dt
œ
dr dt
d) ˆ dr ‰ ˆ d) ‰ dt u) œ dt [(cos ))i (sin ))j] r dt [( sin ))i (cos ))j] Ê dy dx dr d) dr d) dt œ dt cos ) r dt sin ); v † j œ dt and v † j œ dt sin ) r dt cos )
ur r
cos ) r ddt) sin ) Ê
dr dt
dy dt
dy dt
v†i œ
dx dt
and
sin ) r ddt) cos )
dy (b) ur œ (cos ))i (sin ))j Ê v † ur œ dx dt cos ) dt sin ) d) d) ‰ ˆ dr ‰ œ ˆ dr dt cos ) r dt sin ) (cos ) ) dt sin ) r dt cos ) (sin ) ) by part (a),
Ê v † ur œ
dr dt
; therefore,
œ
dr dt
dx dt
cos )
dy dt
sin );
u) œ (sin ))i (cos ))j Ê v † u) œ sin ) dy dt cos ) dr d) dr d) ˆ ‰ ˆ œ dt cos ) r dt sin ) ( sin )) dt sin ) r dt cos )‰ (cos )) by part (a) Ê v † u) œ r ddt) ; dx dt
therefore, r ddt) œ dx dt sin ) 8. r œ f()) Ê
dr dt
œ f w () )
d) dt
Ê
d# r dt#
dy dt
cos ) #
œ f ww ()) ˆ ddt) ‰ f w ())
d# ) dt#
;vœ
dr dt
ur r
d) dt
u) "Î#
"Î#
# d) ‰ r sin ) ddt) ‰ i ˆsin ) dr r# ˆ ddt) ‰ “ œ ’af w b# f # “ dt r cos ) dt j Ê kvk œ Þ ÞÞ Þ ÞÞ d) dr kv ‚ ak œ kx y y xk , where x œ r cos ) and y œ r sin ). Then dx dt œ (r sin )) dt (cos )) dt
œ ˆcos )
‰# ’ˆ dr dt
dr dt
# # d) dr d) ˆ d) ‰# (r sin )) ddt#) (cos )) ddt#r ; dy dt dt (r cos )) dt dt œ (r cos )) dt (sin # # ˆ d) ‰# (r cos )) ddt#) (sin )) ddt#r . Then kv ‚ ak Ê œ (2 cos )) ddt) dr dt (r sin )) dt # $ $ d) d# r d) ˆ dr ‰# œ (after much algebra) r# ˆ ddt) ‰ r ddt#) dr œ ˆ ddt) ‰ Šf 2 f † f ww 2af w b2 ‹ dt r dt dt# 2 dt dt
Ê
d# x dt# d# y dt#
Ê ,œ
œ (2 sin ))
kv ‚a k kv k
œ
dr dt
dr dt
f 2 f†f ww 2af w b2 $Î# af w b# f # ‘
9. (a) Let r œ 2 t and ) œ 3t Ê vœ
))
ˆ ddt) ‰ ;
dr dt
œ 1 and
d) dt
d# r dt#
œ3 Ê #
ur r ddt) u) Ê v(1) œ ur 3u) ; a œ ’ ddt#r
œ
d# ) dt#
# r ˆ ddt) ‰ “ ur
œ 0. The halfway point is (1ß 3) Ê t œ 1; #
’r ddt#) 2 dr dt
d) dt “ u)
Ê a(1) œ 9ur 6u)
(b) It takes the beetle 2 min to crawl to the origin Ê the rod has revolved 6 radians # # Ê L œ '0 É[f())]# cf w ())d# d) œ '0 Ɉ2 3) ‰ ˆ "3 ‰ d) œ '0 É4 6
6
œ '0 É 37 129 ) ) d) œ 6
#
œ È37
" 6
dL dt
4) 3
)# 9
" 9
'06 È() 6)# 1 d) œ "3 ’ ()#6) È() 6)# 1 "# ln ¸) 6 È() 6)# 1¸“ ' !
dL dt
œ ˆ ddtr ‚ mv‰ Šr ‚ m
d# r dt# ‹
Ê
dL dt
œ (v ‚ mv) (r ‚ ma) œ r ‚ ma ; F œ ma Ê krck$ r
œ r ‚ ma œ r ‚ Š krck$ r‹ œ krck$ (r ‚ r) œ 0 Ê L œ constant vector
â â i â 11. (a) ur ‚ u) œ â cos ) â â sin )
j sin ) cos )
â kâ â 0 â œ k Ê a right-handed frame of unit vectors â 0â
œ ( sin ))i (cos ))j œ u) and ddu)) œ ( cos ))i (sin ))j œ ur Þ ÞÞ Þ Þ Þ ÞÞ ÞÞ ÞÞ Þ Þ ÞÞ (c) From Eq. (7), v œ rur r)u) zk Ê a œ v œ a r ur r ur b ˆr )u) r) u) r) u) ‰ z k Þ# ÞÞ ÞÞ ÞÞ ÞÞ œ Š r r) ‹ ur ˆr) 2r )‰ u) z k (b)
d)
ln ŠÈ37 6‹ ¸ 6.5 in.
10. L(t) œ r(t) ‚ mv(t) Ê œ ma Ê
" 3
6
dur d)
12. (a) x œ r cos ) Ê dx œ cos ) dr r sin ) d); y œ r sin ) Ê dy œ sin ) dr r cos ) d); thus dx# œ cos# ) dr# 2r sin ) cos ) dr d) r# sin# ) d)# and Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 13 Additional and Advanced Exercises dy# œ sin# ) dr# 2r sin ) cos ) dr d) r# cos# ) d)# Ê dx# dy# dz# œ dr# r# d)# dz# (b) (c) r œ e) Ê dr œ e) d) Ê L œ '0
ln 8
Èdr# r# d)# dz#
œ '0 Èe#) e#) e#) d) ln 8
œ '0 È3e) d) œ ’È3 e) “ ln 8
ln 8 0
œ 8È3 È3 œ 7È3
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
861
862
Chapter 13 Vector-Valued Functions and Motion in Space
NOTES:
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
CHAPTER 14 PARTIAL DERIVATIVES 14.1 FUNCTIONS OF SEVERAL VARIABLES 1. (a) (b) (c) (d) (e) (f)
Domain: all points in the xy-plane Range: all real numbers level curves are straight lines y x œ c parallel to the line y œ x no boundary points both open and closed unbounded
2. (a) (b) (c) (d)
Domain: set of all (xß y) so that y x 0 Ê y x Range: z 0 level curves are straight lines of the form y x œ c where c 0 boundary is Èy x œ 0 Ê y œ x, a straight line
(e) closed (f) unbounded 3. (a) Domain: all points in the xy-plane (b) Range: z 0 (c) level curves: for f(xß y) œ 0, the origin; for f(xß y) œ c 0, ellipses with center (!ß 0) and major and minor axes along the x- and y-axes, respectively (d) no boundary points (e) both open and closed (f) unbounded 4. (a) Domain: all points in the xy-plane (b) Range: all real numbers (c) level curves: for f(xß y) œ 0, the union of the lines y œ „ x; for f(xß y) œ c Á 0, hyperbolas centered at (0ß 0) with foci on the x-axis if c 0 and on the y-axis if c 0 (d) no boundary points (e) both open and closed (f) unbounded 5. (a) Domain: all points in the xy-plane (b) Range: all real numbers (c) level curves are hyperbolas with the x- and y-axes as asymptotes when f(xß y) Á 0, and the x- and y-axes when f(xß y) œ 0 (d) no boundary points (e) both open and closed (f) unbounded 6. (a) Domain: all (xß y) Á (0ß y) (b) Range: all real numbers (c) level curves: for f(xß y) œ 0, the x-axis minus the origin; for f(xß y) œ c Á 0, the parabolas y œ cx# minus the origin (d) boundary is the line x œ 0
864
Chapter 14 Partial Derivatives
(e) open (f) unbounded 7. (a) Domain: all (xß y) satisfying x# y# 16 (b) Range: z "4 (c) (d) (e) (f)
level curves are circles centered at the origin with radii r 4 boundary is the circle x# y# œ 16 open bounded
8. (a) (b) (c) (d) (e) (f)
Domain: all (xß y) satisfying x# y# Ÿ 9 Range: 0 Ÿ z Ÿ 3 level curves are circles centered at the origin with radii r Ÿ 3 boundary is the circle x# y# œ 9 closed bounded
9. (a) (b) (c) (d) (e) (f)
Domain: (xß y) Á (0ß 0) Range: all real numbers level curves are circles with center (!ß 0) and radii r 0 boundary is the single point (0ß 0) open unbounded
10. (a) (b) (c) (d) (e) (f)
Domain: all points in the xy-plane Range: 0 z Ÿ 1 level curves are the origin itself and the circles with center (0ß 0) and radii r 0 no boundary points both open and closed unbounded
11. (a) Domain: all (xß y) satisfying 1 Ÿ y x Ÿ 1 (b) Range: 1# Ÿ z Ÿ 1# (c) (d) (e) (f)
level curves are straight lines of the form y x œ c where 1 Ÿ c Ÿ 1 boundary is the two straight lines y œ 1 x and y œ 1 x closed unbounded
12. (a) Domain: all (xß y), B Á 0 (b) Range: 1# z 1# (c) (d) (e) (f)
level curves are the straight lines of the form y œ cx, c any real number and x Á 0 boundary is the line x œ 0 open unbounded
13. f
14. e
15. a
16. c
17. d
18. b
Section 14.1 Functions of Several Variables 19. (a)
(b)
20. (a)
(b)
21. (a)
(b)
22. (a)
(b)
865
866
Chapter 14 Partial Derivatives
23. (a)
(b)
24. (a)
(b)
25. (a)
(b)
Section 14.1 Functions of Several Variables 26. (a)
(b)
27. (a)
(b)
28. (a)
(b)
#
#
867
29. f(xß y) œ 16 x# y# and Š2È2ß È2‹ Ê z œ 16 Š2È2‹ ŠÈ2‹ œ 6 Ê 6 œ 16 x# y# Ê x# y# œ 10 30. f(xß y) œ Èx# 1 and (1ß 0) Ê D œ È1# 1 œ 0 Ê x# 1 œ 0 Ê x œ 1 or x œ 1 31. f(xß y) œ 'x
y
1 1 t#
dt at ŠÈ2ß È2‹ Ê z œ tan" y tan" x; at ŠÈ2ß È2‹ Ê z œ tan" È2 tan" ŠÈ2‹
œ 2 tan" È2 Ê tan" y tan" x œ 2 tan" È2
_
n
32. f(xß y) œ ! Š xy ‹ at (1ß 2) Ê z œ
œ
n 0
Ê y œ 2x
" 1 Š xy ‹
œ
y yx
; at (1ß 2) Ê z œ
2 #1
œ2 Ê 2œ
y yx
Ê 2y 2x œ y
868
Chapter 14 Partial Derivatives
33.
34.
35.
36.
37.
38.
39.
40.
41. f(xß yß z) œ Èx y ln z at (3ß 1ß 1) Ê w œ Èx y ln z; at (3ß 1ß 1) Ê w œ È3 (1) ln 1 œ 2 Ê Èx y ln z œ 2
Section 14.1 Functions of Several Variables
869
42. f(xß yß z) œ ln ax# y z# b at ("ß #ß ") Ê w œ ln ax# y z# b ; at ("ß #ß ") Ê w œ ln (1 2 1) œ ln 4 Ê ln 4 œ ln ax# y z# b Ê x# y z# œ 4 _
(x b y)n n! zn
43. g(xß yß z) œ !
nœ0
_
at (ln 2ß ln 4ß 3) Ê w œ !
œ eÐln 8ÑÎ3 œ eln 2 œ 2 Ê 2 œ eÐxyÑÎz Ê 44. g(xß yß z) œ 'x
y
d) È1 )#
'È2 z
dt tÈt# 1
xy z
nœ0
(x b y)n n! zn
œ eÐxyÑÎz ; at (ln 2ß ln 4ß 3) Ê w œ eÐln 2ln 4ÑÎ3
œ ln 2
at ˆ0ß "# ß 2‰ Ê w œ csin" )d x csec" td È2 y
z
œ sin" y sin" x sec" z sec" ŠÈ2‹ Ê w œ sin" y sin" x sec" z Ê w œ sin"
" #
sin" 0 sec" 2
1 4
œ
1 4
Ê
1 #
1 4
; at ˆ0ß "# ß 2‰
œ sin" y sin" x sec" z
45. f(xß yß z) œ xyz and x œ 20 t, y œ t, z œ 20 Ê w œ (20 t)(t)(20) along the line Ê w œ 400t 20t# Ê
dw dt
œ 400 40t;
dw dt
œ 0 Ê 400 40t œ 0 Ê t œ 10 and
d# w dt#
œ 40 for all t Ê yes, maximum at t œ 10
Ê x œ 20 10 œ 10, y œ 10, z œ 20 Ê maximum of f along the line is f(10ß 10ß 20) œ (10)(10)(20) œ 2000 46. f(xß yß z) œ xy z and x œ t 1, y œ t 2, z œ t 7 Ê w œ (t 1)(t 2) (t 7) œ t# 4t 5 along the line Ê
dw dt
œ 2t 4;
dw dt
œ 0 Ê 2t 4 œ 0 Ê t œ 2 and
d# w dt#
œ 2 for all t Ê yes, minimum at t œ 2 Ê x œ 2 1 œ 1,
y œ 2 2 œ 0, and z œ 2 7 œ 9 Ê minimum of f along the line is f(1ß 0ß 9) œ (1)(0) 9 œ 9 ‰ 47. w œ 4 ˆ Th d
"Î#
"Î#
km) œ 4 ’ (290 5K)(16.8 “ K/km
¸ 124.86 km Ê must be
" #
(124.86) ¸ 63 km south of Nantucket
48. The graph of f(x" ß x# ß x$ ß x% ) is a set in a five-dimensional space. It is the set of points (x" ß x# ß x$ ß x% ß f(x" ß x# ß x$ ß x% )) for (x" ß x# ß x$ ß x% ) in the domain of f. The graph of f(x" ß x# ß x$ ß á ß xn ) is a set in an (n 1)-dimensional space. It is the set of points (x" ß x# ß x$ ß á ß xn ß f(x" ß x# ß x$ ß á ß xn )) for (x" ß x# ß x$ ß á ß xn ) in the domain of f. 49-52. Example CAS commands: Maple: with( plots ); f := (x,y) -> x*sin(y/2) + y*sin(2*x); xdomain := x=0..5*Pi; ydomain := y=0..5*Pi; x0,y0 := 3*Pi,3*Pi; plot3d( f(x,y), xdomain, ydomain, axes=boxed, style=patch, shading=zhue, title="#49(a) (Section 14.1)" ); plot3d( f(x,y), xdomain, ydomain, grid=[50,50], axes=boxed, shading=zhue, style=patchcontour, orientation=[-90,0], title="#49(b) (Section 14.1)" ); # (b) L := evalf( f(x0,y0) ); # (c) plot3d( f(x,y), xdomain, ydomain, grid=[50,50], axes=boxed, shading=zhue, style=patchcontour, contours=[L], orientation=[-90,0], title="#49(c) (Section 14.1)" ); 53-56. Example CAS commands: Maple: eq := 4*ln(x^2+y^2+z^2)=1; implicitplot3d( eq, x=-2..2, y=-2..2, z=-2..2, grid=[30,30,30], axes=boxed, title="#53 (Section 14.1)" );
870
Chapter 14 Partial Derivatives
57-60. Example CAS commands: Maple: x := (u,v) -> u*cos(v); y := (u,v) ->u*sin(v); z := (u,v) -> u; plot3d( [x(u,v),y(u,v),z(u,v)], u=0..2, v=0..2*Pi, axes=boxed, style=patchcontour, contours=[($0..4)/2], shading=zhue, title="#57 (Section 14.1)" ); 49-60. Example CAS commands: Mathematica: (assigned functions and bounds will vary) For 49 - 52, the command ContourPlot draws 2-dimensional contours that are z-level curves of surfaces z = f(x,y). Clear[x, y, f] f[x_, y_]:= x Sin[y/2] y Sin[2x] xmin= 0; xmax= 51; ymin= 0; ymax= 51; {x0, y0}={31, 31}; cp= ContourPlot[f[x,y], {x, xmin, xmax}, {y, ymin, ymax}, ContourShading Ä False]; cp0= ContourPlot[[f[x,y], {x, xmin, xmax}, {y, ymin, ymax}, Contours Ä {f[x0,y0]}, ContourShading Ä False, PlotStyle Ä {RGBColor[1,0,0]}]; Show[cp, cp0] For 53 - 56, the command ContourPlot3D will be used and requires loading a package. Write the function f[x, y, z] so that when it is equated to zero, it represents the level surface given. For 53, the problem associated with Log[0] can be avoided by rewriting the function as x2 + y2 +z2 - e1/4 1/x/y; q1 := Int( Int( f(x,y), y=1..x ), x=1..3 ); evalf( q1 ); value( q1 ); evalf( value(q1) ); 71-76. Example CAS commands: Maple: f := (x,y) -> exp(x^2); c,d := 0,1; g1 := y ->2*y; g2 := y -> 4; q5 := Int( Int( f(x,y), x=g1(y)..g2(y) ), y=c..d ); value( q5 ); plot3d( 0, x=g1(y)..g2(y), y=c..d, color=pink, style=patchnogrid, axes=boxed, orientation=[-90,0], scaling=constrained, title="#71 (Section 15.1)" ); r5 := Int( Int( f(x,y), y=0..x/2 ), x=0..2 ) + Int( Int( f(x,y), y=0..1 ), x=2..4 ); value( r5); value( q5-r5 ); 67-76. Example CAS commands: Mathematica: (functions and bounds will vary) You can integrate using the built-in integral signs or with the command Integrate. In the Integrate command, the integration begins with the variable on the right. (In this case, y going from 1 to x).
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
949
950
Chapter 15 Multiple Integrals
Clear[x, y, f] f[x_, y_]:= 1 / (x y) Integrate[f[x, y], {x, 1, 3}, {y, 1, x}] To reverse the order of integration, it is best to first plot the region over which the integration extends. This can be done with ImplicitPlot and all bounds involving both x and y can be plotted. A graphics package must be loaded. Remember to use the double equal sign for the equations of the bounding curves. Clear[x, y, f] t^2; k := t -> 3*t^2; a,b := 0,2; ds := ( D(g)^2 + D(h)^2 + D(k)^2 )^(1/2): 'ds' = ds(t)*'dt'; F := f(g,h,k): 'F(t)' = F(t); Int( f, s=C..NULL ) = Int( simplify(F(t)*ds(t)), t=a..b ); `` = value(rhs(%)); Mathematica: (functions and domains may vary) Clear[x, y, z, r, t, f] f[x_,y_,z_]:= Sqrt[1 30x2 10y]
#
# (a) # (b) # (c)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
32 20
œ 1, 232 45
;
Section 16.2 Vector Fields, Work, Circulation, and Flux 1001 {a,b}= {0, 2}; x[t_]:= t y[t_]:= t2 z[t_]:= 3t2 r[t_]:= {x[t], y[t], z[t]} v[t_]:= D[r[t], t] mag[vector_]:=Sqrt[vector.vector] Integrate[f[x[t],y[t],z[t]] mag[v[t]], {t, a, b}] N[%] 16.2 VECTOR FIELDS, WORK, CIRCULATION, AND FLUX 1. f(xß yß z) œ ax# y# z# b `f `y
"Î#
# $Î#
œ y a x# y # z b
and
similarly,
œ
y x # y # z#
and
`f `z
" #
2. f(xß yß z) œ ln Èx# y# z# œ `f `y
`f `x
Ê
`f `z
$Î#
# $Î#
œ z a x# y # z b
`f `x
ln ax# y# z# b Ê
œ
3. g(xß yß z) œ ez ln ax# y# b Ê
œ "# ax# y# z# b
Ê ™fœ
z x # y # z#
`g `x
`g `y
œ x# 2x y# ,
(2x) œ x ax# y# z# b
Ê ™fœ
œ
" #
$Î#
; similarly,
x i y j z k ax# y# z# b$Î#
Š x# y"# z# ‹ (2x) œ
x x # y # z#
;
x i y j zk x# y# z#
œ x# 2y y# and
`g `z
œ ez
z Ê ™ g œ Š x#2xy# ‹ i Š x# 2y y# ‹ j e k
`g `x
4. g(xß yß z) œ xy yz xz Ê
œ y z,
`g `y
`g `z
œ x z, and
œ y x Ê ™ g œ (y z)i (B z)j (x y)k
5. kFk inversely proportional to the square of the distance from (xß y) to the origin Ê È(M(xß y))# (N(xß y))# œ
k x# y#
, k 0; F points toward the origin Ê F is in the direction of n œ
Ê F œ an , for some constant a 0. Then M(xß y) œ Ê È(M(xß y))# (N(xß y))# œ a Ê a œ
k x# y#
ax È x# y#
Ê Fœ
x È x# y#
and N(xß y) œ
kx ax# y# b$Î#
i
i
y È x# y#
j
ay È x# y#
ky ax# y# b$Î#
j , for any constant k 0
6. Given x# y# œ a# b# , let x œ Èa# b# cos t and y œ Èa# b# sin t. Then r œ ŠÈa# b# cos t‹ i ŠÈa# b# sin t‹ j traces the circle in a clockwise direction as t goes from 0 to 21 Ê v œ ŠÈa# b# sin t‹ i ŠÈa# b# cos t‹ j is tangent to the circle in a clockwise direction. Thus, let F œ v Ê F œ yi xj and F(0ß 0) œ 0 . 7. Substitute the parametric representations for r(t) œ x(t)i y(t)j z(t)k representing each path into the vector field F , and calculate the work W œ 'C F † (a) F œ 3ti 2tj 4tk and
dr dt
(b) F œ 3t# i 2tj 4t% k and œ
7 3
2œ
dr dt
.
œijk Ê F† dr dt
dr dt
œ 9t Ê W œ '0 9t dt œ 1
œ i 2tj 4t$ k Ê F †
dr dt
œ 7t# 16t( Ê W œ '0 a7t# 16t( b dt œ 73 t$ 2t) ‘ ! 1
13 3
(c) r" œ ti tj and r# œ i j tk ; F" œ 3ti 2tj and F# œ 3i 2j 4tk and
d r# dt
œ k Ê F# †
d r# dt
d r" dt
9 #
œ i j Ê F" †
d r" dt
"
œ 5t Ê W" œ '0 5t dt œ 1
œ 4t Ê W# œ '0 4t dt œ 2 Ê W œ W" W# œ 1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
9 #
5 #
;
1002 Chapter 16 Integration in Vector Fields 8. Substitute the parametric representation for r(t) œ x(t)i y(t)j z(t)k representing each path into the vector field F, and calculate the work W œ 'C F †
dr dt
.
" ‰ (a) F œ ˆ t# 1 j and
dr dt
œijk Ê F†
" ‰ (b) F œ ˆ t# 1 j and
dr dt
œ i 2tj 4t$ k Ê F †
dr dt
Ê F# †
d r# dt
œ 0 Ê W œ '0
" t# 1
dt œ
1
œ
2t t# 1 Ê W and ddtr" œ i j
dr dt
" ‰ (c) r" œ ti tj and r# œ i j tk ; F" œ ˆ t# 1 j 1
Ê W œ '0
" t# 1
œ
" t# 1
œ '0
"
dt œ ctan" td ! œ
1
Ê
2t t# 1 F" † ddtr"
1 4 "
dt œ cln at# 1bd ! œ ln 2 œ
" t# 1
; F# œ
" #
j and
œk
d r# dt
1 4
9. Substitute the parametric representation for r(t) œ x(t)i y(t)j z(t)k representing each path into the vector field F, and calculate the work W œ 'C F † (a) F œ Èti 2tj Ètk and (b) F œ t# i 2tj tk and
dr dt
dr dt
dr dt
.
œijk Ê F†
" œ 2Èt 2t Ê W œ '0 ˆ2Èt 2t‰ dt œ 43 t$Î# t# ‘ ! œ 1
dr dt
œ i 2tj 4t$ k Ê F †
dr dt
œ 4t% 3t# Ê W œ '0 a4t% 3t# b dt œ 45 t& t$ ‘ ! œ "5 1
(c) r" œ ti tj and r# œ i j tk ; F" œ 2tj Èt k and œ 1; F# œ Èti 2j k and
d r# dt
œ k Ê F# †
" 3
œ i j Ê F" †
d r" dt
d r" dt
"
œ 2t Ê W" œ '0 2t dt 1
œ 1 Ê W# œ '0 dt œ 1 Ê W œ W" W# œ 0 1
d r# dt
10. Substitute the parametric representation for r(t) œ x(t)i y(t)j z(t)k representing each path into the vector field F, and calculate the work W œ 'C F †
dr dt
. œ 3t# Ê W œ '0 3t# dt œ 1 1
(a) F œ t# i t# j t# k and
dr dt
œijk Ê F†
(b) F œ t$ i t' j t& k and
dr dt
œ i 2tj 4t$ k Ê F †
%
œ ’ t4
t) 4
"
94 t* “ œ !
dr dt
œ t$ 2t( 4t) Ê W œ '0 at$ 2t( 4t) b dt 1
17 18
(c) r" œ ti tj and r# œ i j tk ; F" œ t# i and F# œ i tj tk and
dr dt
d r# dt
œ k Ê F# †
d r# dt
œ i j Ê F" †
d r" dt
œ t Ê W# œ '0 t dt œ 1
d r" dt " #
œ t# Ê W" œ '0 t# dt œ 1
Ê W œ W" W# œ
" 3
;
5 6
11. Substitute the parametric representation for r(t) œ x(t)i y(t)j z(t)k representing each path into the vector field F, and calculate the work W œ 'C F †
dr dt
. œ 3t# 1 Ê W œ '0 a3t# 1b dt œ ct$ td ! œ 2
(a) F œ a3t# 3tb i 3tj k and
†
(b) F œ a3t# 3tb i 3t% j k
Ê F†
dr dt
1
"
dr & $ # dt œ 6t 4t 3t 3t 1 " Ê W œ 0 a6t& 4t$ 3t# 3tb dt œ t' t% t$ 3# t# ‘ ! œ 3# r" œ ti tj and r# œ i j tk ; F" œ a3t# 3tb i k and ddtr" œ i j Ê F" † ddtr" œ 3t# 1 " Ê W" œ 0 a3t# 3tb dt œ t$ 32 t# ‘ ! œ "# ; F# œ 3tj k and ddtr# œ k Ê F# † ddtr# Ê W œ W" W# œ 12
'
(c)
dr dt œ i j k Ê F and ddtr œ i 2tj 4t$ k
'
3t œ 1 Ê W# œ '0 dt œ 1
12. Substitute the parametric representation for r(t) œ x(t)i y(t)j z(t)k representing each path into the vector field F, and calculate the work W œ 'C F † (a) F œ 2ti 2tj 2tk and
dr dt
dr dt
.
œijk Ê F†
(b) F œ at# t% b i at% tb j at t# b k and
dr dt
dr dt
œ 6t Ê W œ '0 6t dt œ c3t# d ! œ 3
œ i 2tj 4t$ k Ê F †
Ê W œ '0 a6t& 5t% 3t# b dt œ ct' t& t$ d ! œ 3 1
1
"
dr dt
œ 6t& 5t% 3t#
"
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
1
Section 16.2 Vector Fields, Work, Circulation, and Flux 1003 (c) r" œ ti tj and r# œ i j tk ; F" œ ti tj 2tk and F# œ (1 t)i (t 1)j 2k and
d r# dt
œ k Ê F# †
dr" dt
œ i j Ê F" †
d r" dt
Ê F†
dr dt
œ 2t$ Ê work œ '0 2t$ dt œ
1
œ 2 Ê W# œ '0 2 dt œ 2 Ê W œ W" W# œ 3 1
d r# dt
13. r œ ti t# j tk , 0 Ÿ t Ÿ 1, and F œ xyi yj yzk Ê F œ t$ i t# j t$ k and 1
œ 2t Ê W" œ '0 2t dt œ ";
œ i 2tj k
dr dt
" #
14. r œ (cos t)i (sin t)j 6t k , 0 Ÿ t Ÿ 21, and F œ 2yi 3xj (x y)k Ê F œ (2 sin t)i (3 cos t)j (cos t sin t)k and œ 3 cos# t 2sin2 t œ 32 t
3 4
" 6
sin 2t t
cos t " 6
sin 2t 2
" 6
dr dt
œ ( sin t)i (cos t)j 6" k Ê F †
sin t Ê work œ '0 ˆ3 cos# t 2 sin2 t
sin t
" 6
#1 t‘ !
cos
21
" 6
cos t
" 6
dr dt
sin t‰ dt
œ1
15. r œ (sin t)i (cos t)j tk , 0 Ÿ t Ÿ 21, and F œ zi xj yk Ê F œ ti (sin t)j (cos t)k and dr dt
œ (cos t)i (sin t)j k Ê F †
œ cos t t sin t
t 2
sin 2t 4
dr dt
œ t cos t sin# t cos t Ê work œ '0 at cos t sin# t cos tb dt 21
#1
sin t‘ ! œ 1
16. r œ (sin t)i (cos t)j 6t k , 0 Ÿ t Ÿ 21, and F œ 6zi y# j 12xk Ê F œ ti acos# tbj (12 sin t)k and dr dt
œ (cos t)i (sin t)j "6 k Ê F †
dr dt
œ t cos t sin t cos# t 2 sin t
Ê work œ '0 at cos t sin t cos# t 2 sin tb dt œ cos t t sin t 21
1 3
#1
cos$ t 2 cos t‘ ! œ 0
17. x œ t and y œ x# œ t# Ê r œ ti t# j , 1 Ÿ t Ÿ 2, and F œ xyi (x y)j Ê F œ t$ i at t# b j and dr dt
œ i 2tj Ê F †
dr dt
œ t$ a2t# 2t$ b œ 3t$ 2t# Ê 'C xy dx (x y) dy œ 'C F †
#
œ 34 t% 23 t$ ‘ " œ ˆ12
16 ‰ 3
ˆ 34 23 ‰ œ
45 4
18 3
œ
dr dt
dt œ 'c" a3t$ 2t# b dt #
69 4
18. Along (0ß 0) to (1ß 0): r œ ti , 0 Ÿ t Ÿ 1, and F œ (x y)i (x y)j Ê F œ ti tj and
dr dt
œi Ê F†
dr dt
œ t;
Along (1ß 0) to (0ß 1): r œ (1 t)i tj , 0 Ÿ t Ÿ 1, and F œ (x y)i (x y)j Ê F œ (1 2t)i j and dr dr dt œ i j Ê F † dt œ 2t; Along (0ß 1) to (0ß 0): r œ (1 t)j , 0 Ÿ t Ÿ 1, and F œ (x y)i (x y)j Ê F œ (t 1)i (1 t)j and dr dt
œ j Ê F †
dr dt
œ t 1 Ê 'C (x y) dx (x y) dy œ '0 t dt '0 2t dt '0 (t 1) dt œ '0 (4t 1) dt 1
1
1
1
"
œ c2t# td ! œ 2 1 œ 1 19. r œ xi yj œ y# i yj , 2 y 1, and F œ x# i yj œ y% i yj Ê Ê
c1
'C F † T ds œ '2
F†
dr dy
dr dt
œ 2yi j and F †
dr dy
œ 2y& y
" 4‰ dy œ '2 a2y& yb dy œ 3" y' #" y# ‘ # œ ˆ 3" #" ‰ ˆ 64 3 # œ
20. r œ (cos t)i (sin t)j , 0 Ÿ t Ÿ Ê F†
c1
dr dy
1 #
, and F œ yi xj Ê F œ (sin t)i (cos t)j and
œ sin# t cos# t œ 1 Ê
'C F † dr œ '0
1Î2
dr dt
3 #
63 3
œ 39 #
œ ( sin t)i (cos t)j
(1) dt œ 1#
21. r œ (i j) t(i 2j) œ (1 t)i (1 2t)j , 0 Ÿ t Ÿ 1, and F œ xyi (y x)j Ê F œ a1 3t 2t# b i tj and dr dt
œ i 2j Ê F †
dr dt
œ 1 5t 2t# Ê work œ 'C F †
dr dt
dt œ '0 a1 5t 2t# b dt œ t 52 t# 23 t$ ‘ ! œ 1
22. r œ (2 cos t)i (2 sin t)j , 0 Ÿ t Ÿ 21, and F œ ™ f œ 2(x y)i 2(x y)j Ê F œ 4(cos t sin t)i 4(cos t sin t)j and ddtr œ (2 sin t)i (2 cos t)j Ê F †
"
dr dt
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
25 6
1004 Chapter 16 Integration in Vector Fields œ 8 asin t cos t sin# tb 8 acos# t cos t sin tb œ 8 acos# t sin# tb œ 8 cos 2t Ê work œ 'C ™ f † dr œ 'C F †
dt œ '0 8 cos 2t dt œ c4 sin 2td #!1 œ 0 21
dr dt
23. (a) r œ (cos t)i (sin t)j , 0 Ÿ t Ÿ 21, F" œ xi yj , and F# œ yi xj Ê F" œ (cos t)i (sin t)j , and F# œ ( sin t)i (cos t)j Ê F" †
dr dt
dr dt
œ ( sin t)i (cos t)j ,
œ 0 and F# †
dr dt
œ sin# t cos# t œ 1
Ê Circ" œ '0 0 dt œ 0 and Circ# œ '0 dt œ 21; n œ (cos t)i (sin t)j Ê F" † n œ cos# t sin# t œ 1 and 21
21
F# † n œ 0 Ê Flux" œ '0 dt œ 21 and Flux# œ '0 0 dt œ 0 21
21
(b) r œ (cos t)i (4 sin t)j , 0 Ÿ t Ÿ 21 Ê F# œ (4 sin t)i (cos t)j Ê F" †
dr dt
dr dt
œ ( sin t)i (4 cos t)j , F" œ (cos t)i (4 sin t)j , and
œ 15 sin t cos t and F# †
dr dt
œ 4 Ê Circ" œ '0 15 sin t cos t dt 21
#1 œ "25 sin# t‘ ! œ 0 and Circ# œ '0 4 dt œ 81; n œ Š È417 cos t‹ i Š È"17 sin t‹ j Ê F" † n 21
œ
4 È17
cos# t
4 È17
sin# t and F# † n œ È1517 sin t cos t Ê Flux" œ '0 (F" † n) kvk dt œ '0 Š È417 ‹ È17 dt 21
21
# ‘ #1 œ 81 and Flux# œ '0 (F# † n) kvk dt œ '0 Š È1517 sin t cos t‹ È17 dt œ 15 2 sin t ! œ 0 21
21
24. r œ (a cos t)i (a sin t)j , 0 Ÿ t Ÿ 21, F" œ 2xi 3yj , and F# œ 2xi (x y)j Ê
œ (a sin t)i (a cos t)j ,
dr dt
F" œ (2a cos t)i (3a sin t)j , and F# œ (2a cos t)i (a cos t a sin t)j Ê n kvk œ (a cos t)i (a sin t)j , F" † n kvk œ 2a# cos# t 3a# sin# t, and F# † n kvk œ 2a# cos# t a# sin t cos t a# sin# t Ê Flux" œ '0 a2a# cos# t 3a# sin# tb dt œ 2a# 2t 21
sin 2t ‘ #1 4 !
3a# 2t
Flux# œ '0 a2a# cos# t a# sin t cos t a# sin# tb dt œ 2a# 2t 21
25. F" œ (a cos t)i (a sin t)j ,
d r" dt
œ (a sin t)i (a cos t)j Ê F" †
sin 2t ‘ #1 4 ! d r" dt
sin 2t ‘ #1 4 !
œ 1a# , and
a# #
#1
csin# td ! a# 2t
sin 2t ‘ #1 4 !
œ 1a#
œ 0 Ê Circ" œ 0; M" œ a cos t,
N" œ a sin t, dx œ a sin t dt, dy œ a cos t dt Ê Flux" œ 'C M" dy N" dx œ '0 aa# cos# t a# sin# tb dt œ '0 a# dt œ a# 1;
1
1
F # œ ti ,
d r# dt
œ i Ê F# †
d r# dt
œ t Ê Circ# œ 'ca t dt œ 0; M# œ t, N# œ 0, dx œ dt, dy œ 0 Ê Flux# a
œ 'C M# dy N# dx œ 'ca 0 dt œ 0; therefore, Circ œ Circ" Circ# œ 0 and Flux œ Flux" Flux# œ a# 1 a
26. F" œ aa# cos# tb i aa# sin# tb j ,
d r" dt
œ (a sin t)i (a cos t)j Ê F" †
d r" dt
œ a$ sin t cos# t a$ cos t sin# t
Ê Circ" œ '0 aa$ sin t cos# t a$ cos t sin# tb dt œ 2a3 ; M" œ a# cos# t, N" œ a# sin# t, dy œ a cos t dt, 1
$
dx œ a sin t dt Ê Flux" œ 'C M" dy N" dx œ '0 aa$ cos$ t a$ sin$ tb dt œ 1
F # œ t# i ,
d r# dt
œ i Ê F# †
d r# dt
œ t# Ê Circ# œ 'ca t# dt œ a
2a$ 3
4 3
a$ ;
; M# œ t# , N# œ 0, dy œ 0, dx œ dt
Ê Flux# œ 'C M# dy N# dx œ 0; therefore, Circ œ Circ" Circ# œ 0 and Flux œ Flux" Flux# œ 27. F" œ (a sin t)i (a cos t)j ,
d r" dt
œ (a sin t)i (a cos t)j Ê F" †
d r" dt
4 3
a$
œ a# sin# t a# cos# t œ a#
Ê Circ" œ '0 a# dt œ a# 1 ; M" œ a sin t, N" œ a cos t, dx œ a sin t dt, dy œ a cos t dt 1
Ê Flux" œ 'C M" dy N" dx œ '0 aa# sin t cos t a# sin t cos tb dt œ 0; F# œ tj , 1
dr# dt
œ i Ê F# †
d r# dt
œ0
Ê Circ# œ 0; M# œ 0, N# œ t, dx œ dt, dy œ 0 Ê Flux# œ 'C M# dy N# dx œ 'ca t dt œ 0; therefore, a
Circ œ Circ" Circ# œ a# 1 and Flux œ Flux" Flux# œ 0
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 16.2 Vector Fields, Work, Circulation, and Flux 1005 28. F" œ aa# sin# tb i aa# cos# tb j ,
d r" dt
œ (a sin t)i (a cos t)j Ê F" †
Ê Circ" œ '0 aa$ sin$ t a$ cos$ tb dt œ 1
4 3
d r" dt
œ a$ sin$ t a$ cos$ t
a$ ; M" œ a# sin# t, N" œ a# cos# t, dy œ a cos t dt, dx œ a sin t dt
Ê Flux" œ 'C M" dy N" dx œ '0 aa$ cos t sin# t a$ sin t cos# tb dt œ 1
2 3
a$ ; F# œ t# j ,
d r# dt
œ i Ê F# †
d r# dt
œ0
Ê Circ# œ 0; M# œ 0, N# œ t# , dy œ 0, dx œ dt Ê Flux# œ 'C M# dy N# dx œ 'ca t# dt œ 23 a$ ; therefore, a
Circ œ Circ" Circ# œ
4 3
a$ and Flux œ Flux" Flux# œ 0
29. (a) r œ (cos t)i (sin t)j , 0 Ÿ t Ÿ 1, and F œ (x y)i ax# y# b j Ê F œ (cos t sin t)i acos# t sin# tb j Ê F †
dr dt
(b)
sin 2t 1 ‘1 4 sin t ! œ # r œ (1 2t)i , 0 Ÿ t Ÿ 1, and F œ (x y)i ax# y# b j Ê ddtr œ 2i and F œ (1 2t)i (1 2t)# j Ê 1 " F † ddtr œ 4t 2 Ê C F † T ds œ 0 (4t 2) dt œ c2t# 2td ! œ 0 r" œ (1 t)i tj , 0 Ÿ t Ÿ 1, and F œ (x y)i ax# y# b j Ê ddtr" œ i j and F œ (1 2t)i a1 2t
'
(c)
œ (sin t)i (cos t)j and
œ sin t cos t sin# t cos t Ê 'C F † T ds
œ '0 a sin t cos t sin# t cos tb dt œ "2 sin# t 1
dr dt
Ê F†
t #
'
œ (2t 1) a1 2t 2t# b œ 2t# Ê Flow" œ 'C F †
d r" dt
"
0 Ÿ t Ÿ 1, and F œ (x y)i ax# y# b j Ê œ i a2t# 2t 1b j Ê F † "
œ t# 23 t$ ‘ ! œ
" 3
dr# dt
dr# dt
dr" dt
œ '0 2t# dt œ 1
2 3
2t# b j
; r# œ ti (t 1)j ,
œ i j and F œ i at# t# 2t 1b j
œ 1 a2t# 2t 1b œ 2t 2t# Ê Flow# œ 'C F †
dr# dt
#
Ê Flow œ Flow" Flow# œ
2 3
" 3
œ '0 a2t 2t# b dt 1
œ1
30. From (1ß 0) to (0ß 1): r" œ (1 t)i tj , 0 Ÿ t Ÿ 1, and F œ (x y)i ax# y# b j Ê
d r" dt
œ i j ,
F œ i a1 2t 2t# b j , and n" kv" k œ i j Ê F † n" kv" k œ 2t 2t# Ê Flux" œ '0 a2t 2t# b dt 1
"
œ t# 23 t$ ‘ ! œ
" 3
;
From (0ß 1) to (1ß 0): r# œ ti (1 t)j , 0 Ÿ t Ÿ 1, and F œ (x y)i ax# y# b j Ê
d r# dt
œ i j ,
#
F œ (1 2t)i a1 2t 2t b j , and n# kv# k œ i j Ê F † n# kv# k œ (2t 1) a1 2t 2t# b œ 2 4t 2t# Ê Flux# œ '0 a2 4t 2t# b dt œ 2t 2t# 23 t$ ‘ ! œ 23 ; 1
"
From (1ß 0) to (1ß 0): r$ œ (1 2t)i , 0 Ÿ t Ÿ 1, and F œ (x y)i ax# y# b j Ê #
d r$ dt
œ 2i ,
#
F œ (1 2t)i a1 4t 4t b j , and n$ kv$ k œ 2j Ê F † n$ kv$ k œ 2 a1 4t 4t b Ê Flux$ œ 2 '0 a1 4t 4t# b dt œ 2 t 2t# 43 t$ ‘ ! œ 1
31. F œ Èx#y y# i
"
x È x# y#
2 3
Ê Flux œ Flux" Flux# Flux$ œ
j on x# y# œ 4;
at (2ß 0), F œ j ; at (0ß 2), F œ i ; at (2ß 0), È F œ j ; at (!ß 2), F œ i ; at ŠÈ2ß È2‹ , F œ #3 i "# j ; at ŠÈ2ß È2‹ , F œ Fœ
È3 #
È3 #
i "# j ; at ŠÈ2ß È2‹ ,
i "# j ; at ŠÈ2ß È2‹ , F œ
È3 #
i "# j
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
" 3
2 3
2 3
œ
" 3
1006 Chapter 16 Integration in Vector Fields 32. F œ xi yj on x# y# œ 1; at (1ß 0), F œ i ; at (1ß 0), F œ i ; at (0ß 1), F œ j ; at (0ß 1), F œ j ; at Š "# ß
È3 # ‹,
Fœ
" #
at Š "# ß
È3 # ‹,
F œ "# i
at Š "# ß
È3 # ‹,
Fœ
at Š "# ß
È3 # ‹,
" #
i
i È3 #
È3 #
È3 #
j;
j;
j;
F œ "# i
È3 #
j.
33. (a) G œ P(xß y)i Q(xß y)j is to have a magnitude Èa# b# and to be tangent to x# y# œ a# b# in a counterclockwise direction. Thus x# y# œ a# b# Ê 2x 2yyw œ 0 Ê yw œ xy is the slope of the tangent line at any point on the circle Ê yw œ ba at (aß b). Let v œ bi aj Ê kvk œ Èa# b# , with v in a counterclockwise direction and tangent to the circle. Then let P(xß y) œ y and Q(xß y) œ x Ê G œ yi xj Ê for (aß b) on x# y# œ a# b# we have G œ bi aj and kGk œ Èa# b# . (b) G œ ˆÈx# y# ‰ F œ ŠÈa# b# ‹ F . 34. (a) From Exercise 33, part a, yi xj is a vector tangent to the circle and pointing in a counterclockwise direction Ê yi xj is a vector tangent to the circle pointing in a clockwise direction Ê G œ Èyxi #xjy# is a unit vector tangent to the circle and pointing in a clockwise direction. (b) G œ F 35. The slope of the line through (xß y) and the origin is pointing away from the origin Ê F œ
xi yj È x# y#
y x
Ê v œ xi yj is a vector parallel to that line and
is the unit vector pointing toward the origin.
36. (a) From Exercise 35, Èxxi #yjy# is a unit vector through (xß y) pointing toward the origin and we want kFk to have magnitude Èx# y# Ê F œ Èx# y# Š Èxxi #yjy# ‹ œ xi yj . (b) We want kFk œ
C È x# y#
37. F œ 4t$ i 8t# j 2k and 38. F œ 12t# j 9t# k and
dr dt
where C Á 0 is a constant Ê F œ dr dt
œ i 2tj Ê F †
œ 3j 4k Ê F †
39. F œ (cos t sin t)i (cos t)k and
dr dt
dr dt
dr dt
40. F œ (2 sin t)i (2 cos t)j 2k and
œ 12t$ Ê Flow œ '0 12t$ dt œ c3t% d ! œ 48 2
#
1
œ ( sin t)i (cos t)k Ê F †
dr dt
yj Š Èxxi #yjy# ‹ œ C Š xx#i y# ‹.
œ 72t# Ê Flow œ '0 72t# dt œ c24t$ d ! œ 24
Ê Flow œ '0 ( sin t cos t 1) dt œ "2 cos# t 1
C È x# y#
1 t‘ !
dr dt
"
œ sin t cos t 1
œ ˆ "# 1‰ ˆ "# 0‰ œ 1
œ (2 sin t)i (2 cos t)j 2k Ê F †
dr dt
œ 4 sin# t 4 cos# t 4 œ 0
Ê Flow œ 0 41. C" : r œ (cos t)i (sin t)j tk , 0 Ÿ t Ÿ Ê F†
dr dt
1 #
Ê F œ (2 cos t)i 2tj (2 sin t)k and
dr dt
œ ( sin t)i (cos t)j k
œ 2 cos t sin t 2t cos t 2 sin t œ sin 2t 2t cos t 2 sin t Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 16.2 Vector Fields, Work, Circulation, and Flux 1007 Ê Flow" œ '0
1Î2
C# : r œ j
1 #
1Î#
( sin 2t 2t cos t 2 sin t) dt œ "2 cos 2t 2t sin t 2 cos t 2 cos t‘ !
(1 t)k , 0 Ÿ t Ÿ 1 Ê F œ 1(1 t)j 2k and
dr dt
Ê Flow# œ '0 1 dt œ c1td "! œ 1;
1 #
œ k Ê F†
dr dt
œ 1 1 ;
œ 1
1
C$ : r œ ti (1 t)j , 0 Ÿ t Ÿ 1 Ê F œ 2ti 2(1 t)k and
dr dt
œij Ê F†
Ê Flow$ œ '0 2t dt œ ct# d ! œ 1 Ê Circulation œ (1 1) 1 1 œ 0 1
42. F †
œx
dr dt
dx dt
y
dr dt
œ 2t
"
dy dt
z
` f dx ` x dt
œ
dz dt
` f dy ` y dt
by the chain rule Ê Circulation œ 'C F †
dr dt
` f dz ` z dt
dt œ 'a
, where f(xß yß z) œ b
d dt afaratbbb
" #
ax# y# x# b Ê F †
dr dt
œ
d dt afaratbbb
dt œ farabbb faraabb. Since C is an entire ellipse,
rabb œ raab, thus the Circulation œ 0. 43. Let x œ t be the parameter Ê y œ x# œ t# and z œ x œ t Ê r œ ti t# j tk , 0 Ÿ t Ÿ 1 from (0ß 0ß 0) to (1ß 1ß 1) Ê œ
dr dt
œ i 2tj k and F œ xyi yj yzk œ t$ i t# j t$ k Ê F †
œ t$ 2t$ t$ œ 2t$ Ê Flow œ '0 2t$ dt 1
" #
44. (a) F œ ™ axy# z$ b Ê F † œ 'a
(b)
dr dt
b
d dt afaratbbb
dr dt
œ
` f dx ` x dt
` f dy ` y dt
` z dz ` z dt
œ
df dt
, where f(xß yß z) œ xy# z$ Ê )C F †
dr dt
dt
dt œ farabbb faraabb œ 0 since C is an entire ellipse.
Ð2ß1ß1Ñ
'C F † ddtr œ 'Ð1ß1ß1Ñ
Ð#ß"ß"Ñ
axy# z$ b dt œ cxy# z$ d Ð"ß"ß"Ñ œ (2)(1)# (1)$ (1)(1)# (1)$ œ 2 1 œ 3
d dt
45. Yes. The work and area have the same numerical value because work œ 'C F † dr œ 'C yi † dr œ 'b [f(t)i] † i a
df dt
j‘ dt
[On the path, y equals f(t)]
œ 'a f(t) dt œ Area under the curve b
46. r œ xi yj œ xi f(x)j Ê from the origin Ê F † Ê
'C
dr dx
œ
F † T ds œ 'C F †
dr dx
dr dx
[because f(t) 0]
œ i f w (x)j ; F œ
kx È x# y#
k†y†f (x) È x# y#
dx œ 'a k b
w
d dx
k È x# y#
œ
(xi yj) has constant magnitude k and points away
kx k†f(x)†f (x) Èx# [f(x)]# w
œk
d dx
Èx# [f(x)]# , by the chain rule
Èx# [f(x)]# dx œ k Èx# [f(x)]# ‘ b a
œ k ˆÈb# [f(b)]# Èa# [f(a)]# ‰ , as claimed. 47-52. Example CAS commands: Maple: with( LinearAlgebra );#47 F := r -> < r[1]*r[2]^6 | 3*r[1]*(r[1]*r[2]^5+2) >; r := t -> < 2*cos(t) | sin(t) >; a,b := 0,2*Pi; dr := map(diff,r(t),t); F(r(t)); q1 := simplify( F(r(t)) . dr ) assuming t::real; q2 := Int( q1, t=a..b ); value( q2 ); Mathematica: (functions and bounds will vary): Exercises 47 and 48 use vectors in 2 dimensions Clear[x, y, t, f, r, v] f[x_, y_]:= {x y6 , 3x (x y5 2)}
# (a) # (b) # (c)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
1008 Chapter 16 Integration in Vector Fields {a, b}={0, 21}; x[t_]:= 2 Cos[t] y[t_]:= Sin[t] r[t_]:={x[t], y[t]} v[t_]:= r'[t] integrand= f[x[t], y[t]] . v[t] //Simplify Integrate[integrand,{t, a, b}] N[%] If the integration takes too long or cannot be done, use NIntegrate to integrate numerically. This is suggested for exercises 49 - 52 that use vectors in 3 dimensions. Be certain to leave spaces between variables to be multiplied. Clear[x, y, z, t, f, r, v] f[x_, y_, z_]:= {y y z Cos[x y z], x2 x z Cos[x y z], z x y Cos[x y z]} {a, b}={0, 21}; x[t_]:= 2 Cos[t] y[t_]:= 3 Sin[t] z[t_]:= 1 r[t_]:={x[t], y[t], z[t]} v[t_]:= r'[t] integrand= f[x[t], y[t],z[t]] . v[t] //Simplify NIntegrate[integrand,{t, a, b}] 16.3 PATH INDEPENDENCE, POTENTIAL FUNCTIONS, AND CONSERVATIVE FIELDS 1.
`P `y
œxœ
2.
`P `y
œ x cos z œ
3.
`P `y
œ 1 Á 1 œ
5.
`N `x
œ0Á1œ
6.
`P `y
œ0œ
7.
`f `x
œ 2x Ê f(xß yß z) œ x# g(yß z) Ê
Ê 8.
`f `x
`f `z
`N `z
`N `z
,
,
`M `z `N `z
œyœ ,
`N `z
`M `y `M `z
`M `z
`P `x
,
`N `x
`M `y
œzœ
œ y cos z œ
`P `x
,
`f `x
`N `x
œ0œ
`P `x
,
`N `x
`M `y
œ ex sin y œ `f `y
œ
`f `z
`M `y
Ê Not Conservative
`g `y
œ 3y Ê g(yß z) œ
h(z) Ê
`f `z
œ 2xe
y2z
w
œx
`f `y œ y2z
3y# #
h(z)
2z# C `g `y
œxz Ê
œ z Ê g(yß z) œ zy h(z)
w
xey2z
`f `y
`g `y
œ xey2z Ê
`g `y
œ 0 Ê f(xß yß z)
w
Ê h (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ xey2z C
h (z) œ 2xe
`f `z
`g `y
h(z) Ê f(xß yß z) œ x#
w
œ y sin z Ê f(xß yß z) œ xy sin z g(yß z) Ê
Ê f(xß yß z) œ xy sin z h(z) Ê
`f `y
3y #
#
3y# #
œ x y h (z) œ x y Ê h (z) œ 0 Ê h(z) œ C Ê f(xß yß z)
œ ey2z Ê f(xß yß z) œ xey2z g(yß z) Ê
œ xy sin z C
œ 1 Á 1 œ
Ê Conservative
œ hw (z) œ 4z Ê h(z) œ 2z# C Ê f(xß yß z) œ x#
y2z
10.
Ê Conservative
4.
œ y z Ê f(xß yß z) œ (y z)x g(yß z) Ê
œ xe
`M `y
Ê Not Conservative
œ (y z)x zy C `f `x
œ sin z œ
Ê Not Conservative
Ê f(xß yß z) œ (y z)x zy h(z) Ê
9.
`N `x
Ê Conservative
œ x sin z w
`g `y
œ x sin z Ê
`g `y
œ 0 Ê g(yß z) œ h(z)
w
œ xy cos z h (z) œ xy cos z Ê h (z) œ 0 Ê h(z) œ C Ê f(xß yß z)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 16.3 Path Independence, Potential Functions, and Conservative Fields 1009 11.
`f `z
œ
Ê f(xß yß z) œ
z y # z#
" #
œ (x ln x x) tan (x y) h(y) Ê f(xß yß z) œ `f `y
Ê
12.
œ
" #
`f `x
œ
œ
y y # z# # #
" #
`g `x œ #
œ
ln x sec# (x y) Ê g(xß y)
ln ay# z b (x ln x x) tan (x y) h(y)
sec# (x y) hw (y) œ sec# (x y)
y y # z#
Ê hw (y) œ 0 Ê h(y) œ C Ê f(xß yß z)
ln ay z b (x ln x x) tan (x y) C y 1 x# y# `g `y
œ
z È 1 y # z#
Ê
`f `z
œ
y È 1 y # z#
hw (z) œ
Ê f(xß yß z) œ tan
"
(xy) sin
`f `z
" z
y È 1 y # z#
`g `y
x 1 x# y#
œ
" z
Ê hw (z) œ
`P `y
`N `z
`M `P `N `M `z œ 0 œ `x , `x œ 0 œ `y `g `f # ` y œ ` y œ 2y Ê g(yß z) œ y
œ0œ
œ 2x Ê f(xß yß z) œ x# g(yß z) Ê
`P `y
14. Let F(xß yß z) œ yzi xzj xyk Ê
œ yz Ê f(xß yß z) œ xyz g(yß z) Ê `f `z
œ xyz h(z) Ê Ð3ß5ß0Ñ
'Ð1ß1ß2Ñ
Ê
`N `z
œxœ
Ê M dx N dy P dz is
,
,
`f `y
`M `z
œyœ `g `y
œ xz
`P `x
,
`N `x
`M `y
œzœ `g `y
œ xz Ê
h(z) Ê f(xß yß z) œ x# y# œ h(z)
'Ð0Ð2ß0ß3ß0ßÑ 6Ñ 2x dx 2y dy 2z dz
œ f(2ß 3ß 6) f(!ß !ß !) œ 2# 3# (6)# œ 49
exact;
z È 1 y # z#
Ê h(z) œ ln kzk C
œ hw (z) œ 2z Ê h(z) œ z# C Ê f(xß yß z) œ x# y# z# C Ê
`f `x
x 1 x# y#
(yz) ln kzk C
13. Let F(xß yß z) œ 2xi 2yj 2zk Ê `f `x
œ
Ê g(yß z) œ sin" (yz) h(z) Ê f(xß yß z) œ tan" (xy) sin" (yz) h(z) "
exact;
`f `y
Ê f(xß yß z) œ tan" (xy) g(yß z) Ê
Ê
Ê
`f `x
ln ay# z# b g(xß y) Ê
Ê M dx N dy P dz is
œ 0 Ê g(yß z) œ h(z) Ê f(xß yß z)
œ xy hw (z) œ xy Ê hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ xyz C
yz dx xz dy xy dz œ f(3ß 5ß 0) f(1ß 1ß 2) œ 0 2 œ 2
15. Let F(xß yß z) œ 2xyi ax# z# b j 2yzk Ê Ê M dx N dy P dz is exact;
`f `x
`P `y
`N `z
œ 2z œ
,
`M `z
œ0œ
`P `x
œ 2xy Ê f(xß yß z) œ x# y g(yß z) Ê
Ê g(yß z) œ yz# h(z) Ê f(xß yß z) œ x# y yz# h(z) Ê
`f `z
,
`N `x
`f `y w
œ 2x œ
œ x#
`g `y
`M `y `g `y
œ x# z# Ê
œ z#
œ 2yz h (z) œ 2yz Ê hw (z) œ 0 Ê h(z) œ C
Ê f(xß yß z) œ x# y yz# C Ê 'Ð0ß0ß0Ñ 2xy dx ax# z# b dy 2yz dz œ f("ß #ß $) f(!ß !ß !) œ 2 2(3)# œ 16 Ð1ß2ß3Ñ
16. Let F(xß yß z) œ 2xi y# j ˆ 1 4 z# ‰ k Ê Ê M dx N dy P dz is exact; Ê f(xß yß z) œ x# œ x#
y$ 3
œ ˆ9
27 3
y$ 3
h(z) Ê
`f `x
`P `y
œ0œ
`N `z
`f `z
,
`N `x
œ0œ `f `y
œ
`M `y
`g `y
$
œ y# Ê g(yß z) œ y3 h(z)
4 1 z#
dz œ f(3ß 3ß 1) f(!ß !ß !)
(! ! 0) œ 1
17. Let F(xß yß z) œ (sin y cos x)i (cos y sin x)j k Ê Ê M dx N dy P dz is exact; œ cos y sin x Ê
`P `x
œ0œ
œ hw (z) œ 1 4 z# Ê h(z) œ 4 tan" z C Ê f(xß yß z)
Ð3ß3ß1Ñ
4†
`M `z
œ 2x Ê f(xß yß z) œ x# g(yß z) Ê
4 tan" z C Ê 'Ð0ß0ß0Ñ 2x dx y# dy 1‰ 4
,
`g `y
`f `x
`P `y
œ0œ
`N `z
,
`M `z
œ0œ
`P `x
,
`N `x
œ cos y cos x œ
œ sin y cos x Ê f(xß yß z) œ sin y sin x g(yß z) Ê
œ 0 Ê g(yß z) œ h(z) Ê f(xß yß z) œ sin y sin x h(z) Ê
Ê f(xß yß z) œ sin y sin x z C Ê
' 100101
Ð ß ß Ñ
Ð ß ß Ñ
`f `z
`f `y
`M `y
œ cos y sin x
`g `y
œ hw (z) œ 1 Ê h(z) œ z C
sin y cos x dx cos y sin x dy dz œ f(0ß 1ß 1) f(1ß !ß !)
œ (0 1) (0 0) œ 1 18. Let F(xß yß z) œ (2 cos y)i Š "y 2x sin y‹ j ˆ "z ‰ k Ê Ê M dx N dy P dz is exact; œ
" y
2x sin y Ê
`g `y
œ
" y
`f `x
`P `y
œ0œ
`N `z
,
`M `z
œ0œ
`P `x
œ 2 cos y Ê f(xß yß z) œ 2x cos y g(yß z) Ê
, `f `y
`N `x
œ 2 sin y œ
œ 2x sin y
Ê g(yß z) œ ln kyk h(z) Ê f(xß yß z) œ 2x cos y ln kyk h(z) Ê
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
`f `z
`M `y
`g `y
œ hw (z) œ
" z
1010 Chapter 16 Integration in Vector Fields Ê h(z) œ ln kzk C Ê f(xß yß z) œ 2x cos y ln kyk ln kzk C
Ê 'Ð0ß2ß1Ñ
Ð1ß1Î2ß2Ñ
2 cos y dx Š "y 2x sin y‹ dy 1 #
œ ˆ2 † 0 ln
" z
dz œ f ˆ1ß 1# ß 2‰ f(!ß #ß ")
ln 2‰ (0 † cos 2 ln 2 ln 1) œ ln #
19. Let F(xß yß z) œ 3x# i Š zy ‹ j (2z ln y)k Ê Ê M dx N dy P dz is exact;
`f `x
`P `y
œ
2z y
1 # `N `z
œ
`M `z
,
œ0œ
`P `x
`f `y
œ 3x# Ê f(xß yß z) œ x$ g(yß z) Ê
Ê f(xß yß z) œ x$ z# ln y h(z) Ê œ x$ z# ln y C Ê 'Ð1ß1ß1Ñ 3x# dx Ð1ß2ß3Ñ
`N `x
,
œ0œ
œ
`g `y
œ
`M `y z# y
Ê g(yß z) œ z# ln y h(z)
`f `z
œ 2z ln y hw (z) œ 2z ln y Ê hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z)
z# y
dy 2z ln y dz œ f(1ß 2ß 3) f("ß "ß ")
œ (1 9 ln 2 C) (1 0 C) œ 9 ln 2 #
`P `y
20. Let F(xß yß z) œ (2x ln y yz)i Š xy xz‹ j (xy)k Ê Ê M dx N dy P dz is exact; x# y
œ
`g `y
xz Ê
`f `x
`N `z
œ x œ
,
`M `z
œ y œ
`P `x
,
`N `x
œ 2x ln y yz Ê f(xß yß z) œ x# ln y xyz g(yß z) Ê `f `z
œ 0 Ê g(yß z) œ h(z) Ê f(xß yß z) œ x# ln y xyz h(z) Ê
œ
2x y
`f `y
œ
zœ x# y
`M `y
xz
`g `y
œ xy hw (z) œ xy Ê hw (z) œ 0
Ê h(z) œ C Ê f(xß yß z) œ x# ln y xyz C Ê 'Ð1ß2ß1Ñ (2x ln y yz) dx Š xy xz‹ dy xy dz Ð2ß1ß1Ñ
#
œ f(2ß 1ß 1) f("ß 2ß 1) œ (4 ln 1 2 C) (ln 2 2 C) œ ln 2 21. Let F(xß yß z) œ Š "y ‹ i Š 1z
x y# ‹ j
Ê M dx N dy P dz is exact; Ê
`g `y
œ
" z
Ê g(yß z) œ
Ê f(xß yß z) œ
x y
y z
ˆ zy# ‰ k Ê
`f `x
" y
œ
C Ê 'Ð1ß1ß1Ñ
Ð2ß2ß2Ñ
" y
œ z"# œ
Ê f(xß yß z) œ
h(z) Ê f(xß yß z) œ
y z
`P `y
x y
dx Š 1z
y z
x y# ‹
x y
`N `z
`M `z
,
`f `y zy#
g(yß z) Ê `f `z
h(z) Ê dy
y z#
`P `x
œ0œ
œ
,
`N `x
œ y1# œ
œ yx#
`g `y
œ
" z
`M `y
x y#
hw (z) œ zy# Ê hw (z) œ 0 Ê h(z) œ C
dz œ f(2ß 2ß 2) f("ß 1ß 1) œ ˆ 2#
2 #
C‰ ˆ 1"
œ0 22. Let F(xß yß z) œ Ê `f `x
`P `y
2xi 2yj 2zk x # y # z#
œ 4yz œ 3%
`N `z
,
`M `z
Šand let 3# œ x# y# z# Ê
œ
2x x# y# z# Ê f(xß yß z) œ Ê `` gy œ 0 Ê g(yß z) œ h(z) w œ x# 2z y# z# Ê h (z) œ 0 Ê
Ð2ß2ß2Ñ
Ê 'Ð1ß1ß1Ñ
`P `x
œ 4xz œ 3%
2x dx 2y dy 2z dz x # y # z#
,
`N `x
œ 4xy œ 3%
`3 `x
`M `y
œ
x 3
`3 `y
,
œ
y 3
`3 `z
,
œ 3z ‹
Ê M dx N dy P dz is exact; `f `y
ln ax# y# z# b g(yß z) Ê
œ
2y x # y # z#
Ê f(xß yß z) œ ln ax# y# z# b h(z) Ê
`g `y
`f `z
œ
œ
2y x # y # z#
hw (z)
2z x # y # z#
h(z) œ C Ê f(xß yß z) œ ln ax# y# z# b C
œ f(2ß 2ß 2) f("ß 1ß 1) œ ln 12 ln 3 œ ln 4
23. r œ (i j k) t(i 2j 2k) œ (1 t)i (1 2t)j (1 2t)k, 0 Ÿ t Ÿ 1 Ê dx œ dt, dy œ 2 dt, dz œ 2 dt Ð2ß3ß1Ñ
Ê 'Ð1ß1ß1Ñ y dx x dy 4 dz œ '0 (2t 1) dt (t 1)(2 dt) 4(2) dt œ '0 (4t 5) dt œ c2t# 5td ! œ 3 1
1
24. r œ t(3j 4k), 0 Ÿ t Ÿ 1 Ê dx œ 0, dy œ 3 dt, dz œ 4 dt Ê
' 000304
Ð ß ß Ñ
Ð ß ß Ñ
"
#
x# dx yz dy Š y# ‹ dz
œ '0 a12t# b (3 dt) Š 9t# ‹ (4 dt) œ '0 54t# dt œ c18t# d ! œ 18 1
25.
`P `y
œ0œ
1
#
`N `z
,
`M `z
œ 2z œ
`P `x
,
`N `x
,
`M `z
"
œ0œ
`M `y
Ê M dx N dy P dz is exact Ê F is conservative
Ê path independence 26.
`P `y
œ ˆÈ
yz x # y # z# ‰
$
œ
`N `z
œ ˆÈ
xz $ x # y # z# ‰
œ
`P `x
,
`N `x
œ ˆÈ
xy x # y # z# ‰
$
œ
`M `y
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
" 1
C‰
Section 16.3 Path Independence, Potential Functions, and Conservative Fields 1011 Ê M dx N dy P dz is exact Ê F is conservative Ê path independence 27.
`P `y `f `x
œ0œ œ
2x y
`N `z
,
œ0œ
Ê f(xß y) œ
Ê f(xß y) œ 28.
`M `z
x# y
" y
`N `z
,
`M `z
#
x y
`P `x
`N `x
,
œ 2x y# œ
`P `x
`f `x
œ e ln y Ê f(xß yß z) œ e ln y g(yß z) Ê
œ
ex y
" y#
Ê gw (y) œ
Ê g(y) œ "y C
1 y ‹
œ cos z œ x
`N `x
#
`P `y
,
1 x# y#
œ yx# gw (y) œ
C Ê F œ ™ Šx œ0œ
Ê F is conservative Ê there exists an f so that F œ ™ f;
#
`f `y
g(y) Ê
`M `y
œ
`M `y
x
Ê F is conservative Ê there exists an f so that F œ ™ f; `f `y
ex y
œ
œ y sin z h(z) Ê f(xß yß z) œ e ln y y sin z h(z) Ê x
`g ex `y œ y `f `z œ y x
`g `y
sin z Ê
œ sin z Ê g(yß z)
w
cos z h (z) œ y cos z Ê hw (z) œ 0
Ê h(z) œ C Ê f(xß yß z) œ ex ln y y sin z C Ê F œ ™ ae ln y y sin zb 29.
`P `y
œ0œ
`N `z
`f `x
œ x# y Ê f(xß yß z) œ
,
`M `z
Ê f(xß yß z) œ œ
" 3
x$ xy
œ0œ
" $ 3 x xy " $ z 3 y ze
(a) work œ 'A F † B
dr dt
`P `x
, " 3 " 3
`N `x
`M `y
œ1œ
Ê F is conservative Ê there exists an f so that F œ ™ f; `f `y
x$ xy g(yß z) Ê
y$ h(z) Ê
œx
`g `y
œ y# x Ê
`f `z
`g `y z
œ y# Ê g(yß z) œ
" 3
y$ h(z)
œ hw (z) œ zez Ê h(z) œ zez e C Ê f(xß yß z) ez C Ê F œ ™ ˆ 3" x$ xy 3" y$ zez ez ‰
dt œ 'A F † dr œ "3 x$ xy 3" y$ zez ez ‘ Ð"ß!ß!Ñ œ ˆ 3" 0 0 e e‰ ˆ 3" 0 0 1‰ B
Ð"ß!ß"Ñ
œ1
(b) work œ 'A F † dr œ "3 x$ xy 3" y$ zez ez ‘ Ð"ß!ß!Ñ œ 1 B
Ð"ß!ß"Ñ
(c) work œ 'A F † dr œ "3 x$ xy 3" y$ zez ez ‘ Ð"ß!ß!Ñ œ 1 B
Ð"ß!ß"Ñ
Note: Since F is conservative, 'A F † dr is independent of the path from (1ß 0ß 0) to (1ß 0ß 1). B
30.
`P `y
œ xeyz xyzeyz cos y œ
that F œ ™ f;
`f `x
`N `z
,
`M `z
œ yeyz œ
`P `x
`N `x
,
œ zeyz œ
œ eyz Ê f(xß yß z) œ xeyz g(yß z) Ê
`f `y
`M `y
Ê F is conservative Ê there exists an f so
œ xzeyz
Ê g(yß z) œ z sin y h(z) Ê f(xß yß z) œ xe z sin y h(z) Ê yz
`g `y
`f `z
œ xzeyz z cos y Ê
`g `y
œ z cos y
w
œ xye sin y h (z) œ xyeyz sin y yz
Ê hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ xeyz z sin y C Ê F œ ™ axeyz z sin yb
(a) work œ 'A F † dr œ cxeyz z sin yd Ð"ß!ß"Ñ B
Ð"ß1Î#ß!Ñ
(b) work œ 'A F † dr œ cxeyz z sin yd Ð"ß!ß"Ñ B
Ð"ß1Î#ß!Ñ
(c) work œ 'A F † dr œ cxeyz z sin yd Ð"ß!ß"Ñ B
Ð"ß1Î#ß!Ñ
œ (1 0) (1 0) œ 0 œ0 œ0
Note: Since F is conservative, 'A F † dr is independent of the path from (1ß 0ß 1) to ˆ1ß 1# ß 0‰ . B
31. (a) F œ ™ ax$ y# b Ê F œ 3x# y# i 2x$ yj ; let C" be the path from (1ß 1) to (0ß 0) Ê x œ t 1 and y œ t 1, 0 Ÿ t Ÿ 1 Ê F œ 3(t 1)# (t 1)# i 2(t 1)$ (t 1)j œ 3(t 1)% i 2(t 1)% j and r" œ (t 1)i (t 1)j Ê dr" œ dt i dt j Ê
'C
"
F † dr" œ '0 c3(t 1)% 2(t 1)% d dt 1
œ '0 5(t 1)% dt œ c(t 1)& d ! œ 1; let C# be the path from (0ß 0) to (1ß 1) Ê x œ t and y œ t, 1
"
0 Ÿ t Ÿ 1 Ê F œ 3t% i 2t% j and r# œ ti tj Ê dr# œ dt i dt j Ê 'C F † dr# œ '0 a3t% 2t% b dt 1
1 œ '0 5t% dt œ 1
Ê 'C F † dr œ 'C F † dr" 'C "
#
#
F † dr# œ 2 Ð1ß1Ñ
(b) Since f(xß y) œ x$ y# is a potential function for F, 'Ð1ß1Ñ F † dr œ f(1ß 1) f(1ß 1) œ 2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
1012 Chapter 16 Integration in Vector Fields 32.
`P `y `f `x
œ0œ
`N `z
`M `z
,
œ0œ
`P `x
,
`N `x
œ 2x sin y œ
œ 2x cos y Ê f(xß yß z) œ x# cos y g(yß z) Ê
Ê f(xß yß z) œ x# cos y h(z) Ê (a) (b) (c) (d)
`M `y
`f `z
Ê F is conservative Ê there exists an f so that F œ ™ f; `f `y
œ x# sin y
`g `y
œ x# sin y Ê
`g `y
œ 0 Ê g(yß z) œ h(z)
œ hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ x# cos y C Ê F œ ™ ax# cos yb
'C 2x cos y dx x# sin y dy œ cx# cos yd Ð!ß"Ñ Ð"ß!Ñ œ 0 1 œ 1 'C 2x cos y dx x# sin y dy œ cx# cos yd Ð"ß!Ñ Ð"ß1Ñ œ 1 (1) œ 2 'C 2x cos y dx x# sin y dy œ cx# cos yd Ð"ß!Ñ Ð"ß!Ñ œ 1 1 œ 0 'C 2x cos y dx x# sin y dy œ cx# cos yd Ð"ß!Ñ Ð"ß!Ñ œ 1 1 œ 0
33. (a) If the differential form is exact, then all x, and
`N `x
œ
`M `y
`P `y
`N `z
œ
Ê 2ay œ cy for all y Ê 2a œ c,
`M `z
œ
`P `x
Ê 2cx œ 2cx for
Ê by œ 2ay for all y Ê b œ 2a and c œ 2a
(b) F œ ™ f Ê the differential form with a œ 1 in part (a) is exact Ê b œ 2 and c œ 2 34. F œ ™ f Ê g(xß yß z) œ 'Ð0ß0ß0Ñ F † dr œ 'Ð0ß0ß0Ñ ™ f † dr œ f(xß yß z) f(0ß 0ß 0) Ê ÐxßyßzÑ
`g `z
œ
`f `z
ÐxßyßzÑ
`g `x
œ
`f `x
0,
`g `y
œ
`f `y
0, and
0 Ê ™ g œ ™ f œ F, as claimed
35. The path will not matter; the work along any path will be the same because the field is conservative. 36. The field is not conservative, for otherwise the work would be the same along C" and C# . 37. Let the coordinates of points A and B be axA , yA , zA b and axB , yB , zB b, respectively. The force F œ ai bj ck is conservative because all the partial derivatives of M, N, and P are zero. Therefore, the potential function is fax, y, zb œ ax by cz C, and the work done by the force in moving a particle along any path from A to B is faBb faAb œ f axB , yB , zB b faxA , yA , zA b œ aaxB byB czB Cb aaxA byA czA Cb Ä œ aaxB xA b bayB yA b cazB zA b œ F † BA 38. (a) Let GmM œ C Ê F œ C ’ `P `y
Ê
3yzC ax# y# z# b&Î#
`f `x
œ
œ
xC ax# y# z# b$Î# yC Ê `` gy œ ax# y# z# b$Î#
some f; œ
œ
`N `z
,
x ax# y# z# b$Î# `M `z
œ
i
y ax# y# z# b$Î#
3xzC ax# y# z# b&Î#
Ê f(xß yß z) œ
œ
,
C ax# y# z# b"Î#
0 Ê g(yß z) œ h(z) Ê
Ê h(z) œ C" Ê f(xß yß z) œ
`P `x
C ax# y# z# b"Î#
`f `z
œ
j
`N `x
œ
z ax# y# z# b$Î# 3xyC ax# y# z# b&Î#
g(yß z) Ê zC ax# y# z# b$Î#
`f `y
k“ `M `y
œ
œ
Ê F œ ™ f for
yC ax# y# z# b$Î#
hw (z) œ
C" . Let C" œ 0 Ê f(xß yß z) œ
zC ax# y# z# b$Î# GmM ax# y# z# b"Î#
`g `y
is a potential
function for F. (b) If s is the distance of (xß yß z) from the origin, then s œ Èx# y# z# . The work done by the gravitational field F is work œ 'P F † dr œ ’ Èx#GmM “ y # z# P#
T#
"
T"
œ
GmM
s#
GmM
s"
œ GmM Š s"#
"
s" ‹ ,
as claimed.
16.4 GREEN'S THEOREM IN THE PLANE 1. M œ y œ a sin t, N œ x œ a cos t, dx œ a sin t dt, dy œ a cos t dt Ê `N `y
`M `x
œ 0,
`M `y
œ 1,
œ 0;
Equation (11):
)C M dy N dx œ '021 [(a sin t)(a cos t) (a cos t)(a sin t)] dt œ '021 0 dt œ 0;
' ' Š ``Mx ``Ny ‹ dx dy œ ' ' 0 dx dy œ 0, Flux R
`N `x
R
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
œ 1, and
Section 16.4 Green's Theorem in the Plane 1013 Equation (12):
)C M dx N dy œ '021 [(a sin t)(a sin t) (a cos t)(a cos t)] dt œ '021 a# dt œ 21a# ; Èa c x
' ' Š ``Nx ``My ‹ dx dy œ ' ' ca cc a
R
#
œ 2a
ˆ 1#
1‰ #
#
#
2 dy dx œ 'ca 4Èa# x# dx œ 4 ’ x2 Èa# x# a
sin" xa “
a
ca
#
œ 2a 1, Circulation
2. M œ y œ a sin t, N œ 0, dx œ a sin t dt, dy œ a cos t dt Ê Equation (11):
a# #
)C M dy N dx œ '0
21
`M `x
`M `y
œ 0,
`N `x
œ 1,
œ 0, and
`N `y
œ 0;
a# sin t cos t dt œ a# "2 sin# t‘ ! œ 0; ' ' 0 dx dy œ 0, Flux #1
R
21 #1 Equation (12): )C M dx N dy œ '0 aa# sin# tb dt œ a# 2t sin4 2t ‘ ! œ 1a# ; ' ' Š ``Nx ``My ‹ dx dy
œ ' ' 1 dx dy œ '0
21
R
'0
a
r dr d) œ '0 21
R
a# #
#
d) œ 1a , Circulation
3. M œ 2x œ 2a cos t, N œ 3y œ 3a sin t, dx œ a sin t dt, dy œ a cos t dt Ê `N `y
`M `x
œ 2,
`M `y
œ 0,
`N `x
œ 0, and
œ 3;
Equation (11):
)C M dy N dx œ '021 [(2a cos t)(a cos t) (3a sin t)(a sin t)] dt
œ '0 a2a# cos# t 3a# sin# tb dt œ 2a# 2t 21
sin 2t ‘ #1 4 !
3a# 2t
sin 2t ‘ #1 4 !
œ 21a# 31a# œ 1a# ;
' ' Š ``Mx ``Ny ‹ œ ' ' 1 dx dy œ ' ' r dr d) œ ' a## d) œ 1a# , Flux 0 0 0 21
R
a
21
R
Equation (12):
)C M dx N dy œ '021 [(2a cos t)(a sin t) (3a sin t)(a cos t)] dt
#1 œ '0 a2a# sin t cos t 3a# sin t cos tb dt œ 5a# 12 sin# t‘ ! œ 0; ' ' 0 dx dy œ 0, Circulation 21
R
4. M œ x# y œ a$ cos# t, N œ xy# œ a$ cos t sin# t, dx œ a sin t dt, dy œ a cos t dt Ê ``Mx œ 2xy, ``My œ x2 , ``Nx œ y# , and ``Ny œ 2xy; Equation (11):
)C M dy N dx œ '021 aa% cos$ t sin t a% cos t sin$ tb œ ’ a4
%
cos% t
a% 4
sin% t“
' ' Š ``Mx ``Ny ‹ dx dy œ ' ' (2xy 2xy) dx dy œ 0, Flux R
#1 !
œ 0;
R
)C M dx N dy œ '021 aa% cos# t sin# t a% cos# t sin# tb dt œ '021 a2a% cos# t sin# tb dt 21 41 %1 œ '0 "# a% sin# 2t dt œ a4 '0 sin# u du œ a4 u2 sin42u ‘ ! œ 1#a ; ' ' Š ``Nx ``My ‹ dx dy œ ' ' ay# x# b dx dy
Equation (12):
%
%
21 a 21 œ '0 '0 r# † r dr d) œ '0 a4
%
5. M œ x y, N œ y x Ê
`M `x
%
R
d) œ
1 a% #
, Circulation
œ 1,
`M `y
œ 1,
`N `x
œ 1,
`N `y
Circ œ ' ' [1 (1)] dx dy œ 0
R
œ 1 Ê Flux œ ' ' 2 dx dy œ '0
1
R
'01 2 dx dy œ 2;
R
6. M œ x# 4y, N œ x y# Ê
`M `x
œ 2x,
`M `y
œ 4,
`N `x
œ 1,
`N `y
œ 2y Ê Flux œ ' ' (2x 2y) dx dy R
1 1 1 1 " " œ '0 '0 (2x 2y) dx dy œ '0 cx# 2xyd ! dy œ '0 (1 2y) dy œ cy y# d ! œ 2; Circ œ ' ' (1 4) dx dy
œ '0
1
'0 3 dx dy œ 3
R
1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
1014 Chapter 16 Integration in Vector Fields `M `x
7. M œ y# x# , N œ x# y# Ê
`M `y
œ 2x,
`N `x
œ 2y,
œ 2x,
œ 2y Ê Flux œ ' ' (2x 2y) dx dy
`N `y
R
3 x 3 $ œ '0 '0 (2x 2y) dy dx œ '0 a2x# x# b dx œ "3 x$ ‘ ! œ 9; Circ œ ' ' (2x 2y) dx dy
R
3 x 3 œ '0 '0 (2x 2y) dy dx œ '0 x# dx œ 9
`M `x
8. M œ x y, N œ ax# y# b Ê œ '0
1
`M `y
œ 1,
'0 (1 2y) dy dx œ '0 ax x b dx œ x
1
œ 2x,
œ 2y Ê Flux œ ' ' (1 2y) dx dy
`N `y
; Circ œ ' ' (2x 1) dx dy œ '0
1
" 6
#
`N `x
œ 1,
R
œ '0 a2x# xb dx œ 76
R
'0 (2x 1) dy dx x
1
`M `x
9. M œ x ex sin y, N œ x ex cos y Ê
Ècos 2)
1Î4
Ê Flux œ ' ' dx dy œ 'c1Î4 '0 R
œ 1 ex sin y, Î
`M `y
œ ex cos y,
1 4
1Î%
Ècos 2)
1Î4
R
R
y x
, N œ ln ax# y# b Ê
Ê Flux œ ' ' Š x#yy# R
Circ œ ' ' Š x# 2x y#
x x# y# ‹
R
`M `x
11. M œ xy, N œ y# Ê œ '0 Š 3x# 1
#
3x% # ‹
2y x# y# ‹
dx œ
`M `x
dx dy œ '0
1
dx dy œ '0
1
`M `y
œ y,
y x# y#
œ
'1
`M `y
œ
œ 0,
`M `y
œ 0,
Ê Flux œ ' ' (x sin y) dx dy œ '0
1Î2
R
œ
2x x# y#
,
`N `y
œ
" #
2y x# y#
1
`N `y
œ 2y Ê Flux œ ' ' (y 2y) dy dx œ '0
1
R
`M `x
`N `x
Î
1 4
#
1
12. M œ sin y, N œ x cos y Ê
,
;
r dr d) œ '1Î4 ˆ "# cos 2)‰ d) œ
'12 ˆ r sinr ) ‰ r dr d) œ '01 sin ) d) œ 2;
; Circ œ ' ' x dy dx œ '0
" 5
x x# y#
" #
œ ex sin y
)‰ ˆ r cos r dr d) œ '0 cos ) d) œ 0 r#
2
`N `x
œ x,
,
`N `y
œ 1 ex cos y,
r dr d) œ '1Î4 ˆ "# cos 2)‰ d) œ 4" sin 2)‘ 1Î% œ
Circ œ ' ' a1 ex cos y ex cos yb dx dy œ ' ' dx dy œ 'c1Î4 '0 10. M œ tan"
`N `x
œ cos y,
'0
1Î2
R
'xx 3y dy dx #
'xx x dy dx œ '01 ax# x$ b dx œ 1"# #
`N `x
`N `y
œ cos y,
œ x sin y
(x sin y) dx dy œ '0 Š 18 sin y‹ dy œ 18 ; 1Î2
#
#
1Î# Circ œ ' ' [cos y ( cos y)] dx dy œ '0 '0 2 cos y dx dy œ '0 1 cos y dy œ c1 sin yd ! œ 1 1Î2
1Î2
1Î2
R
13. M œ 3xy
x 1 y#
`M `x
, N œ ex tan " y Ê
Ê Flux œ ' ' Š3y R
" 1 y#
œ 3y R
$
ex y
Ê
3 c x#
`M `y
`N `y
œ
" 1 y# 21
œ '0 a$ (1 cos ))$ (sin )) d) œ ’ a4 (1 cos ))% “ 14. M œ y ex ln y, N œ
,
dx dy œ ' ' 3y dx dy œ '0
" 1 y# ‹
21
" 1 y#
œ1
ex y
,
`N `x
œ
ex y
#1 !
'0aÐ1 cos Ñ )
(3r sin )) r dr d)
œ 4a$ a4a$ b œ 0
Ê Circ œ ' ' ’ ey Š1 x
R
ex y ‹“
œ 'c1 'x% b 1 dy dx œ 'c1 ca3 x b ax 1bd dx œ 'c1 ax x 2b dx œ 1
15. M œ 2xy$ , N œ 4x# y# Ê œ '0
1
'0
x$
2xy dy dx œ ' #
1
#
`M `y
œ 6xy# ,
`N `x
"!
2 33
1
2 0 3
x
dx œ
%
1
%
#
dx dy œ ' ' (1) dx dy R
44 15
œ 8xy# Ê work œ )C 2xy$ dx 4x# y# dy œ ' ' a8xy# 6xy# b dx dy R
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 16.4 Green's Theorem in the Plane 1015 `M `y
16. M œ 4x 2y, N œ 2x 4y Ê
œ 2,
`N `x
œ 2 Ê work œ )C (4x 2y) dx (2x 4y) dy
œ ' ' [2 (2)] dx dy œ 4 ' ' dx dy œ 4(Area of the circle) œ 4(1 † 4) œ 161 R
R `M `y
17. M œ y# , N œ x# Ê œ '0
1
1cx
'0
œ 2y,
œ 2x Ê )C y# dx x# dy œ ' ' (2x 2y) dy dx
`N `x
R
(2x 2y) dy dx œ '0 a3x 4x 1b dx œ cx 2x# xd ! œ 1 2 1 œ 0 1
`M `y
18. M œ 3y, N œ 2x Ê
œ 3,
#
`N `x
œ 2 Ê )C 3y dx 2x dy œ ' ' (2 3) dx dy œ '0
`M `y
œ 6,
1
R
œ '0 sin x dx œ 2 1
19. M œ 6y x, N œ y 2x Ê
"
$
`N `x
'0sin x 1 dy dx
œ 2 Ê )C (6y x) dx (y 2x) dy œ ' ' (2 6) dy dx R
œ 4(Area of the circle) œ 161 20. M œ 2x y# , N œ 2xy 3y Ê
`M `y
œ 2y,
`N `x
œ 2y Ê
)C a2x y# b dx (2xy 3y) dy œ ' ' (2y 2y) dx dy œ 0 R
21. M œ x œ a cos t, N œ y œ a sin t Ê dx œ a sin t dt, dy œ a cos t dt Ê Area œ œ
'0
21
" #
aa# cos# t a# sin# tb dt œ
" #
'0
21
'021 aab cos# t ab sin# tb dt œ "# '021 ab dt œ 1ab
" #
)C
" #
)C x dy y dx
x dy y dx
a# dt œ 1a#
22. M œ x œ a cos t, N œ y œ b sin t Ê dx œ a sin t dt, dy œ b cos t dt Ê Area œ œ
" #
" #
23. M œ x œ a cos$ t, N œ y œ sin$ t Ê dx œ 3 cos# t sin t dt, dy œ 3 sin# t cos t dt Ê Area œ
'0
21
œ
" #
œ
3 16
a3 sin# t cos# tb acos# t sin# tb dt œ
u2
sin 2u ‘ %1 4 !
" #
'0
21
a3 sin# t cos# tb dt œ
3 8
œ
3 8
1
24. M œ x œ t# , N œ y œ
t$ 3
t Ê dx œ 2t dt, dy œ at# 1b dt Ê Area œ
È
È
'0
21
" #
" #
'cÈ33 ’t# at# 1b Š t3 t‹ (2t)“ dt œ "# 'cÈ33 ˆ 3" t% t# ‰ dt œ 12 151 t&
œ
8 5
È3
25. (a) M œ f(x), N œ g(y) Ê
`M `y
œ 0,
`N `x
R
(b) M œ ky, N œ hx Ê
`M `y
œ k,
`N `x
È$
31 t$ ‘ È$ œ
œ 0 Ê )C f(x) dx g(y) dy œ ' ' Š ``Nx R
œ ' ' 0 dx dy œ 0
œ h Ê )C ky dx hx dy œ ' ' Š ``Nx
œ ' ' (h k) dx dy œ (h k)(Area of the region)
3 16
'0
sin# u du
)C x dy y dx
œ
$
sin# 2t dt œ
)C x dy y dx
41
R
`M `y ‹
`M `y ‹
" 15
Š9È3 15È3‹
dx dy
dx dy
R
26. M œ xy# , N œ x# y 2x Ê
`M `y
œ 2xy,
`N `x
œ 2xy 2 Ê )C xy# dx ax# y 2xb dy œ ' ' Š ``Nx
œ ' ' (2xy 2 2xy) dx dy œ 2 ' ' dx dy œ 2 times the area of the square R
R
R
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
`M `y ‹
dx dy
1016 Chapter 16 Integration in Vector Fields 27. The integral is 0 for any simple closed plane curve C. The reasoning: By the tangential form of Green's Theorem, with M œ 4x$ y and N œ x% , )C 4x$ y dx x% dy œ ' ' ’ ``x ax% b
` `y
R
œ ' ' ðóóñóóò a4x$ 4x$ b dx dy œ 0.
a4x$ yb“ dx dy
R
0 28. The integral is 0 for any simple closed curve C. The reasoning: By the normal form of Green's theorem, with ` ` $ $ M œ x$ and N œ y$ , )C y$ dy x$ dx œ ' ' ”ðñò ` x ay b ï ` y ax b • dx dy œ 0.
R
0 `M `x
29. Let M œ x and N œ 0 Ê
œ 1 and
`N `y
œ0 Ê
0
)C M dy N dx œ ' ' Š ``Mx ``Ny ‹ dx dy
œ ' ' (1 0) dx dy Ê Area of R œ ' ' dx dy œ )C x dy; similarly, M œ y and N œ 0 Ê R
`N `x
R
œ 0 Ê )C M dx N dy œ ' ' Š ``Nx R
œ ' ' dx dy œ Area of R
Ê )C x dy
R
`M `y ‹
`M `y
œ 1 and
dy dx Ê )C y dx œ ' ' (0 1) dy dx Ê )C y dx R
R
30.
'ab f(x) dx œ Area of R œ )C y dx, from Exercise 29
31. Let $ (xß y) œ 1 Ê x œ
My M
' ' x $ (xßy) dA
œ 'R '
$ (xßy) dA
' ' x dA
œ 'R '
R
dA
' ' x dA
œ
Ê Ax œ ' ' x dA œ ' ' (x 0) dx dy
R
A
R
R
R
œ )C x#
dy, Ax œ ' ' x dA œ ' ' (0 x) dx dy œ ) xy dx, and Ax œ ' ' x dA œ ' ' ˆ 23 x 3" x‰ dx dy
œ)
#
#
" C 3
R
R
" 3
" #
x dy xy dx Ê
C
)C x
dy œ )C xy dx œ
#
" 3
)C x
dy xy dx œ Ax
32. If $ (xß y) œ 1, then Iy œ ' ' x# $ (xß y) dA œ ' ' x# dA œ ' ' ax# 0b dy dx œ R
R
R
R
#
R
" 3
)C
x$ dy,
' ' x# dA œ ' ' a0 x# b dy dx œ ) x# y dx, and ' ' x# dA œ ' ' ˆ 34 x# "4 x# ‰ dy dx C R
R
œ)
" C 4
33. M œ
`f `y
" 4
$
#
x dy x y dx œ , N œ `` xf Ê
`M `y
" 4
œ
)C x ` #f ` y#
,
R
$
#
dy x y dx Ê
`N `x
" 3
œ `` xf# Ê )C #
)C x
`f `y
R
$
dy œ )C x# y dx œ
dx
`f `x
" 4
)C
dy œ ' ' Š `` xf# #
R
x$ dy x# y dx œ Iy
` #f ` y# ‹
dx dy œ 0 for such
curves C 34. M œ
" 4
x# y 3" y$ , N œ x Ê
the ellipse
" 4
`M `y
œ
1 4
x# y# ,
`N `x
œ 1 Ê Curl œ
`N `x
`M `y
œ 1 ˆ "4 x# y# ‰ 0 in the interior of
x# y# œ 1 Ê work œ 'C F † dr œ ' ' ˆ1 4" x# y# ‰ dx dy will be maximized on the region R
R œ {(xß y) | curl F} 0 or over the region enclosed by 1 œ 2y 35. (a) ™ f œ Š x# 2x y# ‹ i Š x# y# ‹ j Ê M œ
2x x# y#
,Nœ
" 4
x# y#
2y x# y#
; since M, N are discontinuous at (0ß 0), we
compute 'C ™ f † n ds directly since Green's Theorem does not apply. Let x œ a cos t, y œ a sin t Ê dx œ a sin t dt, dy œ a cos t dt, M œ
2 a
cos t, N œ
2 a
sin t, 0 Ÿ t Ÿ 21, so 'C ™ f † n ds œ 'C M dy N dx
œ '0 ˆ 2a cos t‰aa cos tb ˆ 2a sin t‰aa sin tb ‘dt œ '0 2acos2 t sin2 tbdt œ 41. Note that this holds for any 21
21
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 16.4 Green's Theorem in the Plane 1017 a 0, so 'C ™ f † n ds œ 41 for any circle C centered at a0, 0b traversed counterclockwise and 'C ™ f † n ds œ 41 if C is traversed clockwise.
(b) If K does not enclose the point (0ß 0) we may apply Green's Theorem: 'C ™ f † n ds œ 'C M dy N dx œ ' ' Š ``Mx R
`N `y ‹
dx dy œ ' ' Š ax2 y2 b2 2 ˆy 2 x 2 ‰
R
2 ˆx 2 y 2 ‰ ‹ ax 2 y 2 b 2
dx dy œ ' ' 0 dx dy œ 0. If K does enclose the point R
(0ß 0) we proceed as in Example 6: Choose a small enough so that the circle C centered at (0ß 0) of radius a lies entirely within K. Green's Theorem applies to the region R that lies between K and C. Thus, as before, 0 œ ' ' Š ``Mx R
`N `y ‹
dx dy
œ 'K M dy N dx 'C M dy N dx where K is traversed counterclockwise and C is traversed clockwise.
Hence by part (a) 0 œ ’ ' M dy N dx “ 41 Ê 41 œ K
'K ™ f † n ds œ œ 0
'K M dy N dx
œ 'K ™ f † n ds. We have shown:
if (0ß 0) lies inside K if (0ß 0) lies outside K
41
36. Assume a particle has a closed trajectory in R and let C" be the path Ê C" encloses a simply connected region R" Ê C" is a simple closed curve. Then the flux over R" is )C F † n ds œ 0, since the velocity vectors F are "
tangent to C" . But 0 œ )C F † n ds œ )C M dy N dx œ ' ' Š ``Mx "
"
R"
`N `y ‹
dx dy Ê Mx Ny œ 0, which is a
contradiction. Therefore, C" cannot be a closed trajectory. 37.
'gg yy
#Ð Ñ
"Ð Ñ
`N `x
'cd 'gg yy ˆ ``Nx dx‰ dy œ 'cd [N(g# (y)ß y) N(g" (y)ß y)] dy #Ð Ñ
dx dy œ N(g# (y)ß y) N(g" (y)ß y) Ê
"Ð Ñ
œ 'c N(g# (y)ß y) dy 'c N(g" (y)ß y) dy œ 'c N(g# (y)ß y) dy 'd N(g" (y)ß y) dy œ 'C N dy 'C N dy d
38.
d
c
#
"
œ )C dy
Ê
'ab 'cd
dy dx œ 'a [M(xß d) M(xß c)] dx œ 'a M(xß d) dx 'a M(xß c) dx œ 'C M dx 'C M dx.
`M `y
)C N dy œ ' '
d
R
`N `x
dx dy
b
b
b
3
Because x is constant along C# and C% , 'C M dx œ 'C M dx œ 0 Ê
Š'C
"
M dx 'C
#
M dx 'C
$
#
M dx 'C
%
%
M dx‹ œ )C M dx Ê 'a
b
'cd
`M `y
"
dy dx œ )C M dx.
39. The curl of a conservative two-dimensional field is zero. The reasoning: A two-dimensional field F œ Mi Nj can be considered to be the restriction to the xy-plane of a three-dimensional field whose k component is zero, and whose i and j components are independent of z. For such a field to be conservative, we must have `N `M `N `M ` x œ ` y by the component test in Section 16.3 Ê curl F œ ` x ` y œ 0. 40. Green's theorem tells us that the circulation of a conservative two-dimensional field around any simple closed curve in the xy-plane is zero. The reasoning: For a conservative field F œ Mi Nj , we have ``Nx œ ``My (component test for conservative fields, Section 16.3, Eq. (2)), so curl F œ
`N `x
`M `y
œ 0. By Green's theorem,
the counterclockwise circulation around a simple closed plane curve C must equal the integral of curl F over the region R enclosed by C. Since curl F œ 0, the latter integral is zero and, therefore, so is the circulation.
The circulation )C F † T ds is the same as the work )C F † dr done by F around C, so our observation that circulation of a conservative two-dimensional field is zero agrees with the fact that the work done by a conservative field around a closed curve is always 0.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
1018 Chapter 16 Integration in Vector Fields 41-44. Example CAS commands: Maple: with( plots );#41 M := (x,y) -> 2*x-y; N := (x,y) -> x+3*y; C := x^2 + 4*y^2 = 4; implicitplot( C, x=-2..2, y=-2..2, scaling=constrained, title="#41(a) (Section 16.4)" ); curlF_k := D[1](N) - D[2](M): # (b) 'curlF_k' = curlF_k(x,y); top,bot := solve( C, y ); # (c) left,right := -2, 2; q1 := Int( Int( curlF_k(x,y), y=bot..top ), x=left..right ); value( q1 ); Mathematica: (functions and bounds will vary) The ImplicitPlot command will be useful for 41 and 42, but is not needed for 43 and 44. In 44, the equation of the line from (0, 4) to (2, 0) must be determined first. Clear[x, y, f]
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